url
stringlengths 7
1.6k
⌀ | fetch_time
int64 1,368,854,001B
1,726,893,573B
⌀ | content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
⌀ | warc_record_offset
int32 35.5k
1.73B
⌀ | warc_record_length
int32 630
783k
⌀ | text
stringlengths 77
589k
| token_count
int32 40
312k
⌀ | char_count
int32 77
526k
⌀ | metadata
stringlengths 439
444
⌀ | score
float64 3.5
5.13
⌀ | int_score
int64 4
5
⌀ | crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.07
1
⌀ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://forums.babypips.com/t/a-surprisingly-noob-question-are-100-pips-worth-1-of-a-currency-move/71312 | 1,713,937,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00694.warc.gz | 227,195,089 | 6,053 | # A surprisingly 'noob' question: are 100 pips worth 1% of a currency move?
I will explain!
I was listening to PowerTrading Radio the other day, and in hearing about the EUR/USD drop from May 2014
described as a “25%” drop, I wondered if I understood this concept at all: if the EUR/USD dropped 2,500 pips from its
1.40 high of last year, then we would talk about a 25% drop; if, however, we were describing what actually
happened, namely a 3,500-pip drop, then should we not call it a 35% drop?
I know that it is not so easy, because the value of pips for each currency pair is not fixed, although if someone
could explain how that applies to the pip-to-percentage question I put above, I would be grateful.
Thank you - and my apologies if this topic has been covered before (I tried a Google search but I could not
find the very direct answer that I was after).
Regards
PipMeHappy
Take current price and multiple by .25.
That’s 25%.
So 1.4 x .25 = .35
So if price went from 1.4 to 1.05 that would be a 25% drop.
I would assume that because any currency pair, or any asset for that matter, can appreciate in value to a ‘unlimited value’ then the percentage change is always calculated from either the maximum value obtained since inception of the asset, or the maximum value obtained within a predetermined time frame - such as the past 12 months.
I’ve never looked at pips in terms of “pip-to-percentage”, however to answer your question, I would assume 100pips are only equal to 1% when the currency pair has had an all time high value of 10,000 pips since inception (being measured from the origin of ‘0’, where the X and Y axis cross) , hence 100 pips IS 1% of 10,000.
To make it simple, if price moves from 1.40 (the determined high) to perhaps 1.10 (the determined low) I would simply say it’s fallen by 21.43%, irrespective of pip value. As long as you can determine the high and the low, the rest is easy.
…or just read matergunner, I obviously speak to much crap
Thank you, Mastergunner and Jezzode, you answered my question perfectly!
I also did want to ask, what is 25% in relation to? Jezzode: you answered that too!
Thank you - I will actually not use this for trading and stick to pips, but it was only to make sense of it.
Cheers!
1.4 - 1.05 = 0.35
0.35 * 100 / 1.4 = 25%
Simple math…
Yes, rindoan, that’s it!
I have not touched algebra since school! It is surprising how you can forget basic stuff like percentage operations.
Cheers.
F
In formula terms:
[(A - B) * 100] / A = C
where A= high, B= low, and C the final percentage differential.
Correct.
1.40 = 14,000 pips.
3500/14,000 = .25 (25%).
100 pips would be worth 1% of a change only from 10,000 pips which is parity. From 1.40 (14,000 pips), 100 pips is 71 basis points (0.71%).
Of course we trading low lifes don’t care about that because we use leverage and pyramiding to make a move of 3500 pips worth 100% of our account. | 802 | 2,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-18 | latest | en | 0.965861 |
https://www.mersenneforum.org/showthread.php?s=7993ac59947b6e963b883f7545931283&t=20868&page=6 | 1,618,811,754,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00118.warc.gz | 990,165,744 | 13,502 | mersenneforum.org Prime Gap Theory
Register FAQ Search Today's Posts Mark Forums Read
2016-11-18, 15:35 #56 robert44444uk Jun 2003 Oxford, UK 22·3·7·23 Posts A new idea Let X = A*P#/(multiple of some primes) We know that the interval X +/- 2P has very few values that might be prime. What is to stop us finding Y = B*Q#/(multiple of some other primes) that is close in value to X, in fact very close to or equal to X+2(P+Q) or X-2(P+Q) Then we would have an interval approximately twice as long as I that would have very few values that might be prime, assuming P is similar in size to Q. By extension we could find Z=C#/(multiple of yet other primes) close to the range so that the interval is 3 times as long, etc. Simple algebra can find Y,Z I think, or am I missing something fundamental here.
2016-11-18, 16:38 #57
science_man_88
"Forget I exist"
Jul 2009
Dumbassville
20C016 Posts
Quote:
Originally Posted by robert44444uk Let X = A*P#/(multiple of some primes) We know that the interval X +/- 2P has very few values that might be prime. What is to stop us finding Y = B*Q#/(multiple of some other primes) that is close in value to X, in fact very close to or equal to X+2(P+Q) or X-2(P+Q) Then we would have an interval approximately twice as long as I that would have very few values that might be prime, assuming P is similar in size to Q. By extension we could find Z=C#/(multiple of yet other primes) close to the range so that the interval is 3 times as long, etc. Simple algebra can find Y,Z I think, or am I missing something fundamental here.
my guess is it probably depends on gcd and a lot more to figure it out to an exact value for Y or Z.
2016-11-22, 11:43 #58
robert44444uk
Jun 2003
Oxford, UK
111100011002 Posts
Quote:
Let X = A*P#/(multiple of some primes) We know that the interval X +/- 2P has very few values that might be prime. What is to stop us finding Y = B*Q#/(multiple of some other primes) that is close in value to X, in fact very close to or equal to X+2(P+Q) or X-2(P+Q) Then we would have an interval approximately twice as long as I that would have very few values that might be prime, assuming P is similar in size to Q. By extension we could find Z=C#/(multiple of yet other primes) close to the range so that the interval is 3 times as long, etc. Simple algebra can find Y,Z I think, or am I missing something fundamental here.
I don't think the hypothesis I outlined above can be the case. The two values X and Y presumably share a large number of factors, given that P# and Q# are closely related, and hence the closest that X and Y would come would be defined by the multiple of their common factors.
Is this right?
Last fiddled with by robert44444uk on 2016-11-22 at 11:46
2017-03-21, 17:15 #59 robert44444uk Jun 2003 Oxford, UK 36148 Posts I am wondering how we might categorise divisors, in order to find those that are most likely to provide records. I'm looking at an approach using two measures, A - number of gaps >10 over a given range for a divisor B - "persistence" - the ratio of the number of gaps of size x+1 to the number of gaps of size x over a given test range for a divisor For a given divisor D, The test range I am using takes in multiple Y and P in Y*P#/D (centre points of large gaps). The specific ranges I am first looking at are D from 1 to 10,000 (squarefree only of course), Y from 1 to 5,000 and P from 97 to 229, so 130,000 tests for each D. I am not looking for all merit 10 gaps, as I am setting the delta =4. The probability of large gaps should be correlated to A and B, with B dominating, as explained below. The persistence measure is exponential. A 60% persistence suggests that 1 in 27,251 gaps of merit >10 will be a gap of merit 30 and 1 in 165 for a 20 merit gap. A 50% persistence on the other hand, 1 in every 1,048,576 gaps of merit >10 will actually be merit 30, and 1 in every 1,024 merit 20. So although the divisor 2 produces possibly the highest number of size 10 gaps (A=0.352%) its persistence ratio looks to be only in the 35% range, suggesting a conversion ratio of 1 in 1,314,132,370 to achieve a 30 merit gap and 1 in 36,251 for a merit 20. Other small Ds are showing >50% persistence, and A in excess of 0.2%. I know that my favoured divisor 46410 is closer to persistence ratio=60% with an A value relatively negligible. The plan is to find a divisor with persistence of 60% and a much higher A value. Grateful for your views. Last fiddled with by robert44444uk on 2017-03-21 at 17:15
2017-03-23, 18:23 #60 mart_r Dec 2008 you know...around... 32·71 Posts To get a persistence of more than 60%, the value for P must be higher. For P around 20,000 you get B>60% with D a primorial >= 17#, without having to cope with a small A value. For P around 50,000, you can even choose, say, D=41#, and still get a decent A~4% (maybe more) with B>60~65%. Downside is, the tests take longer... If you're looking for merit >30, D=30 is most effective for the P's you're looking at.
2017-03-23, 19:21 #61
danaj
"Dana Jacobsen"
Feb 2011
Bangkok, TH
22·227 Posts
Quote:
Originally Posted by mart_r If you're looking for merit >30, D=30 is most effective for the P's you're looking at.
Looking at the top-20 records:
7 D=30
7 D=210
2 D=2310
4 other (3 are maximal gaps, 1 is D=7230)
Looking at allgaps.dat for merits >= 30,
476 D=30
427 D=210
56 D=2310
20 D=6
9 D=46410
This is heavily impacted by what's being searched for, but I believe D=30 has much more searching than other divisors. For all gaps >= 10 merits, D=30 has over 4 times as many results as the next (D=210).
For gaps under 40k, percent of merits that are over 30. Again impacted by search ranges but maybe it tells us something:
7.17% D=30
7.54% D=210
7.99% D=2310
4.55% D=6
4.23% D=46410
27.78% D=9570
2017-03-23, 22:50 #62
robert44444uk
Jun 2003
Oxford, UK
22·3·7·23 Posts
Quote:
Originally Posted by mart_r To get a persistence of more than 60%, the value for P must be higher. For P around 20,000 you get B>60% with D a primorial >= 17#, without having to cope with a small A value. For P around 50,000, you can even choose, say, D=41#, and still get a decent A~4% (maybe more) with B>60~65%. Downside is, the tests take longer... If you're looking for merit >30, D=30 is most effective for the P's you're looking at.
Sorry Mart_r do you have your Ps and D's mixed up? The way I defined it, P is the prime in the primorial, D is the divisor and Y the multiplier.
2017-03-28, 18:49 #63
mart_r
Dec 2008
you know...around...
32×71 Posts
Quote:
Originally Posted by robert44444uk Sorry Mart_r do you have your Ps and D's mixed up? The way I defined it, P is the prime in the primorial, D is the divisor and Y the multiplier.
No, the P's and D's in my post are all where they have to be.
Only my results may vary a bit more or less compared to yours. I'm basically looking at the count of coprimes mod P#. In my example, 50000#/41#, i.e. D=304250263527210, showed a persistance of >60%, according to your definition.
2017-07-19, 22:27 #64 ATH Einyen Dec 2003 Denmark 17·181 Posts Regarding the "new" gap formula by Maynard, Tao and others https://www.youtube.com/watch?v=BH1GMGDYndo https://arxiv.org/abs/1408.4505 G(X) >= C * log X * loglog X * loglogloglog X / (logloglog X)^2 They found out that C can get arbitrarily large as X->infinity. I was curious about the value of this "Maynard-Tao" constant for known gaps: C= Gn * (logloglog Pn)^2 / (log Pn * loglog Pn * loglogloglog Pn) It seems to follow the "Cramér–Shanks–Granville ratio" somewhat and is largest at the smaller gaps. Code: Gn Merit Gn/(ln(Pn)^2) Maynard-Tao Pn 15900 39.62015365 0.09872683 36.37716367 1.936933265397289504398811903696*10^174 18306 38.06696007 0.07915948 34.09959273 7.041097148478282668812106731813*10^208 10716 36.85828850 0.12677617 35.50074155 1.839377720243795270729953508768*10^126 13692 36.59018324 0.09778276 33.93062260 3.254185929142547441117000456865*10^162 26892 36.42056789 0.04932537 30.92865059 4.696226774889053656642126142794*10^320 66520 35.42445941 0.01886489 27.27312299 3.292808201042179724620296543360*10^815 1476 35.31030807 0.84472754 59.57000119 1425172824437699411 1442 34.97568651 0.84833471 59.49933905 804212830686677669 1454 34.11893253 0.80062005 56.90602177 3219107182492871783 1370 33.76518602 0.83218087 58.00949419 418032645936712127 1132 32.28254764 0.92063859 61.34832684 1693182318746371 6582144 13.18288411 0.00002640 7.04128546 8.465069837806447347636518542879*10^216840 5103138 10.22031845 0.00002047 5.45890046 7.695421151871542659687327631743*10^216848 Last fiddled with by ATH on 2017-07-19 at 22:30
2017-07-20, 13:39 #65
CRGreathouse
Aug 2006
10111011000012 Posts
Quote:
Originally Posted by ATH I was curious about the value of this "Maynard-Tao" constant for known gaps: C= Gn * (logloglog Pn)^2 / (log Pn * loglog Pn * loglogloglog Pn) It seems to follow the "Cramér–Shanks–Granville ratio" somewhat and is largest at the smaller gaps.
It's much larger than the Cramér-Shanks-Granville ratio -- which we expect to be bounded, or 'nearly' bounded, unlike the one you mention (Ford-Green-Konyagin-Tao).
The state of the art today is Ford-Green-Konyagin-Maynard-Tao:
C1 = Gn * (log log log Pn) / (log Pn * log log Pn * log log log log Pn).
2017-07-20, 17:10 #66 robert44444uk Jun 2003 Oxford, UK 36148 Posts Using Antonio's examples and the FGKMT formula provides C as follows: Code: Gn C 15900 20.31243206 18306 18.72974166 10716 20.45427849 13692 19.07131193 26892 16.38392518 66520 13.501963 1476 45.22507308 1442 45.29864017 1454 43.03407437 1370 44.3097939 1132 48.34468117 6582144 2.735318749 5103138 2.12060946 Now I have to get my head around what this means I think we can say that we are nowhere near getting maximal gaps outside of the range we are searching suggesting that the merit is going to increase a lot. Last fiddled with by robert44444uk on 2017-07-20 at 17:13
Similar Threads Thread Thread Starter Forum Replies Last Post Nick Number Theory Discussion Group 6 2016-10-14 19:38 firejuggler Math 0 2016-07-11 23:09 Ricie Miscellaneous Math 24 2009-08-14 15:31 jasong Math 3 2005-05-15 04:01 clowns789 Miscellaneous Math 5 2004-01-08 17:09
All times are UTC. The time now is 05:55.
Mon Apr 19 05:55:54 UTC 2021 up 11 days, 36 mins, 0 users, load averages: 1.23, 1.36, 1.41 | 3,313 | 10,350 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-17 | latest | en | 0.942368 |
https://www.tumblr.com/search/n%2014 | 1,511,239,737,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806316.80/warc/CC-MAIN-20171121040104-20171121060104-00070.warc.gz | 904,777,644 | 90,167 | # n 14
So, Voltron has EIGHT Seasons, at the very least.
8 seasons. seasons= 7 letters. 8 + 7 = 15. And we are meant to have 5 paladins. 15 + 5 = 20
If we follow the letters in the word ‘paladin’ and each letter corresponds to a number (A=1, B=2, etc.)
P=16, A=1, L=12, A=1, D=4, I=9, N=14
16+1+12+1+4+9+14=57
And if we add these to the previous number… 57+20= 77
Coran and Allura were in that pod for 10,000 years. If we divide 10,000 by 77, we get 129.87 aka 130.
If we subtract the 7 main characters, as well as Zarkon and Haggar- which makes 9 characters in total- from the 130 we have, we will get 121.
As you can guess, a phrase can be made from those numbers, if we follow the sequence accordingly.
K=11, L=12, A=1, N=14, C=3, E=5
I=9, S=19
C=3, A=1, N=14, O=15, N=14
11+12+1+14+3+5+9+19+3+1+14+15+14 = 121
## so what i’m saying is
4
GUYS THE NUMBERS FOR THE CASE FILE, IF YOU CORRESPOND WITH THE ALPHABET
S=19
A=1
T=20
A=1
N=14
i’m screaming
Numerology: Calculating Your Life Lesson Number
## Numerology Basics
Each one of us has four personal numbers that carry a specific vibration. They are the Life Lesson Number, the Soul Number, the Outer Personality Number, and the Path of Destiny Number. In this post, we’ll be focusing on the Life Lesson number.
The Life Lesson Number represents lessons one must learn in this lifetime and is most significant to career choices. It is derived from your full birthdate.
Though this post will be focused on calculating one’s number using the numbers in your birthday, I’m including information on how to calculate the numeric vibration of letters as well, because it’s intrinsically related.
To work with the value of letters beyond “I”, which is 9, we reduce the value to a single digit. For example, “O” is 15, so we reduce the number by adding the first digit to the second: 1+5=6. The number 15 is then written as 15/6. The letter “T” is 20. We reduce the 20 by adding the first digit to the second: 2+0=2. The number 20 is then written as 20/2.
Number Values of the Alphabet
A 1 J 10/1 S 19/1
B 2 K 11/2 T 20/2
C 3 L 12/3 U 21/3
D 4 M 13/4 V 22/4
E 5 N 14/5 W 23/5
F 6 O 15/6 X 24/6
G 7 P 16/7 Y 25/7
H 8 Q 17/8 Z 26/8
I 9 R 18/9
We can reduce any word to a numeric vibration by using the above table. Before we move on to calculating one’s Life Lesson number, lets experiment with a few words:
3 + 9 + 6 + 5 = 23/5
C R O W
We reduce 23 by adding 2+3=5, which is written 23/5
5 + 9 + 2 + 3 + 8 = 21/3
W I T C H
We reduce 21 by adding 2+1=3, which is written 21/3
3 + 9 + 7 + 8 + 2 = 29/11
L I G H T
We reduce 29 by adding 2+9=11, which is written 29/11
There are four double digits which generally are not reduced. These are referred to as master numbers. Master numbers offer more opportunity for expression, yet demand more effort from the person or thing in question. The master numbers are 11, 22, 33, and 44. Whenever you arrive at one of these numbers in your calculations; retain the master value.
Because the word “light” vibrates to master number 11, it’s written as 29/11 instead of 29/2, which would be the case if reduced entirely. The number 2 is still acknowledged as the base number as the word will fluctuate between the vibrations of 2 and 11.
An individual will fluctuate between any master number and its base digit because the vibration of the master number is too intense to live under continuously. Living under the vibration of one’s base number offers respite while preparing to work with the master number vibration again.
## Calculating Your Life Lesson Number
To calculate your Life Lesson Number, add all the digits of your birthdate together and reduce the sum. Here is an example of how someone with the birthdate of 24 October 1948 would calculate their Life Lesson Number:
10 + 24 + 1948
10 + 24 + 22 (1 + 9 + 4 + 8)
10 + 24 + 22 = 56/11
Since the reduced sum of their birthday is a master number, the sum is not reduced further.
Now that you know how to calculate your Life Lesson Number, read on to see the description of each numeric vibration. All number descriptions come directly from Numerology and The Divine Triangle (1979) by Faith and Dusty Bunker, the reference material for this post.
## Life Lesson Number Descriptions
1: You must learn to be original, strong willing, creative, and innovative. You should have the courage and drive to go ahead into new fields of expression and be a pioneer. You should always go forward, never turn back. At times you may be dictatorial and stubborn because you do not like to be restricted or directed. You are a good executive and work best alone. You are usually efficient and well organized. You are not naturally domestic but can manage well in any situation. You usually like sports and athletics and enjoy the thrill of winning. You are sophisticated, not emotionally romantic, and always appear at the head of social and commercial groups. By learning the lessons of number 1, you become intimately familiar with the universal energy; that probing, seeking, independent spark that moves all creation. You are creative on the physical plane because your pioneer spirit precedes all others and expresses your unique individuality.
2: You are here to learn to become a good mixer. You are a good salesperson, more persuasive than forceful. You should be a support for those in leadership roles, help them to find their goals in life, and remain behind the scenes if necessary. This quality can be a help to you in business because those who benefit from your talents will in turn help you make use of your abilities. In partnerships and groups, you will encounter the lessons you came to earn in this lifetime. Success is then very possible. You must have consideration for others and should bring people together for a common cause. Various professions are open to you as you learn to be adaptable to most things that need to be done. You could select a career in finance, music, medicine, religion, or statistical analysis and research.
3: You are best in intellectual, artistic, or creative endeavors. You need to be expressive to manifest and to see the results of your work. Beauty, fruitfulness, luxury, and pleasure are your keywords. You should have ambition and pride. You must become conscious of the law, and by being an excellent disciplinarian, you will achieve a position of authority over others. 3 combines the daring of 1 with the caution of 2. It is a number of self-expression and freedom. You must guard against becoming a jack-of-all-trades; rather, you should specialize. You could then be successful in artistic, religious, or inventive pursuits. You should avoid routine work because you dislike restriction. You should work alone for the best results. Business partnerships become too disciplined for your freedom-loving nature. You could write, lecture, teach, or find your niche in journalism. Whatever you decide to specialize in, you must use your creative and inspirational talents.
4: You must build a solid foundation on which to base your life. This demands a well ordered system of conduct and morals. Administration or some sort of management would be the best type of employment for you. You want your home life to conform to the culture in which you live. You will provide well for those within your care, and you expect them to respond with respect and dignity. You should become a diligent worker and honestly earn your success. By being thrifty, you will have an adequate savings account as security against any possible losses. You should learn not to take a chance unless it is a sure bet. You should seek high goals. You might want to achieve concrete results quickly, and therefore should strive for patience and perseverance. Learn to face reality and base all your efforts on sound practical reasoning.
5: Your keyword is freedom. If you have “free rein”, you can accomplish wonders, but if you feel bound or limited, you lose your enthusiasm and accomplish little. You would be a good explorer or Peace Corps volunteer, as you learn well by travel and experience. You are a diligent student if interested in the subject, but you may fail in subjects for which you see no useful ends. You should be eager for new experience, and shun monotony. In your quest for knowledge, you will become interested in discovering answers in books and magazines. An avid reader, a fluent talker, and a versatile doer, you are the witty conversationalist and brighten any group by your mere presence. You are here to learn and experience the value of freedom and should not tie yourself down too severely. Your talents once learned, prepare you for a literary career or a a position in sales and dealing with the public.
6: You are here to learn a sense of responsibility for your family and community. 6 is the love and domestic vibration and requires that you be responsive to the social needs of others. A fine sense of balance must be acquired so that you can equalize injustices. This keen sense bestows artistic abilities as well as judgmental talents which can be utilized in the legal system. You should develop compassion and understanding necessary to ease the burdens of those who will naturally be drawn to you. You are among those who serve, teach, and bring comfort to humanity. A wide choice of professions is yours including nursing, teaching, welfare work, ministry, medicine, restaurant enterprises, a legal profession, and possibly veterinary or animal husbandry. You may also choose to end a career in the arts, interior decorating, or hairdressing.
8: This is the number of power and ambition, the number of the executive, the boss, who lives by brain and brawn. You will learn to work and will want to see everyone else working. You can push people to become successful in their own right. You should lead and show by example how to profit in business. You are here to learn to handle power, authority, and money. You can build a business empire and should work to that end. You want success for your family and for the family name as a matter of pride. You want your offspring to carry your name with honor and dignity. Sports is another field open to you, as this number vibration bestows great strength and endurance. Many famous athletes operate under an 8.
9: You should be the universal lover of humanity, patient, kind, and understanding. You are at the peak of life’s expression and must turn and show others the way. You seem to receive wisdom from above; thus you know that the true way of happiness is in service to others. You are the marrying type, strong in passion and compassion. You easily acquire money or wealth, and know how to preserve it. You are never petty, but deal in broad concepts and can attain success in the face of difficulties. You are here to show others the way, through your breadth of thinking. You can choose from many professions; education and medicine are the most usual. You may become an orator, writer, or lecture with equal ease. Communication, foreign service, statesmanship, and leadership positions are easily within your capacity.
11: The keywords here are altruism and community. You came into a unique and testing incarnation. You must practice “love thy neighbor as thyself” and use it as your foundation. Your strong intuitions are of value in gaining wisdom and inspiration. 11 is one of the most difficult vibrations because the demand for high standards is constant. You must learn patience and at the same time be able to make quick decisions. Seek for balance between the material, physical life which has to be considered, and the inspiration, spiritual life which underlies your self-understanding. You can succeed in the field of science because all new inventions and discoveries such as laser rays, research in fields of anti-gravity or kirlian photography, or any area of electronics would appeal to you. You could choose to be an astronomer or an astrologer, or a bible researcher and interpreter. You may become a teacher or writer in the field of philosophy. You are original and creative and could become and inspirational speaker. 11 is an esoteric master number of spiritual import. It bestows courage, power, and talent with strong feelings of leadership. You must not let this power go to your head, since frame and recognition are likely; instead realize that true mastership is service.
22: You must express a basic building urge, accomplish things in a big way, and work with large groups or business concerns. You would enjoy the import-export business which could demand long distance travel and meetings with persons of authority. You like to take an inspirational idea and put it to practical use. Self-knowledge is very valuable to you. 22 gives the promise of success. You know how to use your ability to adjust the physical laws of life and living to demonstrate esoteric wisdom. You could become an executive in banking or financial affairs in a national capacity, or help organize businesses for others as an efficiency expert or the like. As an ambassador to foreign countries, you would demonstrate statesmanship. You like to be occupied in some large enterprise to challenge your power to achieve. Your lesson is to learn to take charge of large organizations and corporations and to handle money efficiently and usefully for the benefit of large groups of people.
33: You should be steady and reliable and develop a strong desire to protect others. You would like to live close to nature, and this urge may influence you to choose a life in agriculture. Your goal would be to produce food on a large scale to provide sustenance for the hungry of the world. You would never be found in any profession that could act destructively to humanity. Your talent may lie along the line of the arts: music to bring harmony, painting to bring beauty, or literature to promote education. Service in the field of medicine an healing could also attract you. Possibly, you would choose the law as a way to protect others through justice. Since the 33 consciousness is almost beyond that of humanity, a place within ministry or priesthood could lead you to the realm of your dreams as world savior. You may be required to sacrifice your own desires for the need of others in order to fulfill your Life Lesson vibration.
44: This number stands for strength and complete mental control over your life while on earth. It requires discipline in every department of life so that you may be instrumental in promoting the material advancement of the world. Your mind must be trained to let the higher forces work within it, and you must keep your body and environment in order so that you are ready for any opportunity to achieve this same results from others. Your high energy potential is meant to further evolution by helping others set their world in order. You should try to promote better ethics and justice in the world of business. You must recognize reality, then use what you learn to alleviate the physical burdens of others. You are the instrument by which this alteration takes place. By displaying bravery, resourcefulness, courage, and discipline; you serve as an example for others. Edgar Cayce is an example of the 44 vibration.
Please remember that these are just general descriptions of the numeric vibrations pertaining to the Life Lesson Number. Not everything may resonate with you, and that is perfectly fine. Numerology is meant to be tool to guide, not a stringent set of rules to live by.
Stay tuned for follow up posts on the Soul Number, the Outer Personality Number, and the Path of Destiny Number
((source))
so i redrew my fave scene from consequences because felix has a tiny waist and locus is gent l e
4
9
N°14 in the Jibcon 2017 edit spam - Sunday Panels: Jared 🦁
## the M A R K of O X I N
created by 14 y/o @amazingphil (+ played by @danisnotonfire )
If you ever wanna remake that game… I’d happily do the art for it ;D
But seriously, that video and game was just so lovely to watch… Dan excitedly playing it and Phil reminiscing was really cute!
Thank you for sharing it with us!
phan doodles | all art | retweet
redbubble shop | commissions (open)
10
NCT in Busan
The next time BTS is on American News
• Interviewer: welcome Rat Monster, Gin, Sugar, I-Hop, Bee, Jimmy, and Jungcook!
• BTS: ...
• ARMY: ....
• BTS: ...so when are we going back to Korea?
You’ve Been Warned
This was written for @thing-you-do-with-that-thing SPN Hiatus Writing Challenge Week #14 (14 already????) and the prompt is “I never wanted to hurt you.”
Master List
Characters: Sam Winchester, Reader, Dean Winchester
It was quiet in the bunker without Dean around. He had gone off to help a hunter friend of his, (female of course) take care of a Rugaru case in Texas. You and Sam didn’t have a case at the moment, so you took the chance to hang out and recharge.
Dean was going to be gone for at least a week, so the two of you planned on watching movies, doing research, and sorting through the bunkers many rooms of treasures. You loved spending time with Sam. If circumstances were different….
You had a long-standing policy not to get involved with other hunters. Mixing business and pleasure never seemed to work out well for you. Especially since you not only worked with Sam, but you lived with him too. It was a recipe for freaking disaster.
But damn if Sam didn’t make it difficult. Those eyes, those dimples, and lord that ass! Not to mention his towering height, his muscles, and his hair. (Not that you were looking or anything !) But then roll in his kindness, his empathy, and his determination to always do the right thing, and it all made up for one very hard to resist guy.
2
no one tells you how to mourn.
All Black Everything (M)
Author’s Note: i have seen the devil and his name is kim jongin. someone get me a fan. this is a continuation of the universe for Did You See? you can read both separately, however the impact is a little stronger at the end if you read the original story first.
Pairing: Kai x Reader (oc; female)
Genre: smut
Rating: NC-17
Warnings: explicit sex; explicit language
Word Count: 3,881
Nini[10:03 PM]: i can’t stop thinking about the other night…
Y/N[10:05 PM]: which night? lmao i’ve seen you every night this week
Nini[10:08 PM]: don’t be like that, duchess. you know exactly which night i’m talking about.
Y/N[10:09 PM]: no, nini.
Y/N[10:09 PM]: youll have to be specific.
Nini[10:09 PM]: tuesday
Y/N[10:10 PM]: what happened tuesday? we did a lot of things that night~
Nini[10:13 PM]: are you gonna make me say it?
Y/N[10:14 PM]: yes.
3
2
Growing up, what was your favorite TV show/star and did they make you want to become an actor?
4
P O K E M O N G E N E R A T I O N S
DRAYDEN — Episode 14 - The Frozen World
Jealous Texts
Anon requests: Can you do one where the reader is dating Jughead but they’re trying to keep their relationship low-key. Reggie flirts with the reader and bby juggie just gets a little insecure and it ends in fluff and cuddles and fluff did I mention fluff?
Description: Texts exchanged between Jughead and (Y/N) over the course of a week
Warnings: none
Word count: 311
A/N: I’ve never written in this format before, so let me know if you like it!
Wednesday, October 12
2:14 pm
To: (Y/N)
Still on for Pop’s tonight?
2:15 pm
Of course Jug
Friday, October 14
6:34 pm
To: (Y/N)
Why was Reggie talking to you?
7:12 pm
Why does it matter?
7:13 pm
To: (Y/N)
Because I worry
8:01 pm
Don’t
8:02 pm
To: (Y/N)
But I do
He wasn’t flirting with you, was he?
8:49 pm
To: (Y/N)
(Y/N)?
9:07 pm
To: (Y/N)
9:36 pm
To: (Y/N)
9:48 pm
To: (Y/N)
(Y/N) I’m sorry
10:14 pm
Not mad. Just a bit frustrated
Talk to you tomorrow
10:15 pm
To: (Y/N)
I’m sorry. I love you
10:18 pm
Love you too
Saturday, October 15
9:56 am
To: (Y/N)
Pop’s?
10:04 am
Yes. Noon.
10:05 am
To: (Y/N)
Great
1:47 pm
To: (Y/N)
I miss you and I love you
1:50 pm
I just left silly. I love you too
Monday, October 17
8:03 am
Just a reminder that I love you :)
8:05 am
To: (Y/N)
:)
Tuesday, October 18
10:13 am
To: (Y/N)
He’s doing it again
10:17 am
Reggie’s just a friend, Jug
10:19 am
To: (Y/N)
Yeah but does he know that?
10:21 am
I’ll make sure he does
10:30 am
To: (Y/N)
Ok
Wednesday, October 19
1:02 pm
1:14 pm
To: (Y/N)
Why would I be mad at you?
1:16 pm
Because you’re not talking to me
1:19 pm
To: (Y/N)
I’m sorry
There’s a lot on my mind
1:22 pm
My house.
9
Be there
1:25 pm
To: (Y/N)
Ok
11:03 pm
To: (Y/N)
Thank you for that cuddle session
I love you
11:04 pm
I love you too
The Language Of Love // Carlos De Vil Imagine
a/n: *has 14 requests to get to but writes Carlos for fun instead*
Your knuckles rapped at the door for what seemed like 5 minutes before finally it opened, revealing Carlos’s beaming face.
“What’d you get?” you asked excitedly, practically bouncing with anticipation.
Without giving you a response, Carlos turned around and walked over to the drawer next to his bed. You followed him, shutting the door behind you.
Carlos pulled a piece of paper out of the drawer and handed it to you quickly.
You read the paper frantically and yelled when you saw the grade on the top.
“A+! Oh my god Carlos you did better than I did!” You yelled excitedly as you pulled him into a tight hug.
The smile did not leave his face as he happily accepted the hug and wrapped his arms around you. | 5,378 | 21,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-47 | latest | en | 0.857976 |
https://fr.slideserve.com/hawa/chapter-6-5758428-powerpoint-ppt-presentation | 1,685,558,358,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00486.warc.gz | 313,084,012 | 19,460 | # CHAPTER 6
Télécharger la présentation
## CHAPTER 6
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. CHAPTER 6 Algebra: Equations and Inequalities
2. 6.3 Applications of Linear Equations
3. Objectives Use linear equations to solve problems. Solve a formula for a variable.
4. Strategy for Solving Word Problems Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the quantities in the problem. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x. Step 3 Write an equation in x that models the verbal conditions of the problem. Step 4 Solve the equation and answer the problem’s question. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.
5. Algebraic Translations of English Phrases
6. Example 1: Education Pays Off This graph shows the average yearly earnings in the United States by highest educational attainment.The average yearly salary of a man with an associate degree exceeds that of a man with some college by \$3 thousand. The average yearly salary of a man with a bachelor’s degree or more exceeds that of a man with some college by \$41 thousand. Combined, three men with these educational attainments earn \$188 thousand. Find the average yearly salary of men with each of these levels of education.
7. Example 2: continued Step 1: Let x represent one of the unknown quantities. Let x = the average yearly salary of a man with some college. Step 2: Represent the other unknown quantities in terms of x. x + 3 = the average yearly salary of a man with an associate degree x + 41 = the average yearly salary of a man with a bachelor’s degree or more.
8. Example 2: continued Step 3: Write an equation in x that models the conditions. x + (x + 3) + (x + 41) = 188 Step 4: Solve the equation and answer the question.
9. Example 2: continued The average salary with some college = 48 The average salary with an associate degree = x + 3 = 48 + 3 = 51 The average salary with a bachelor’s degree or more = x + 41 = 48 + 41 = 89. Some college = \$48 thousand per year Associate degree = \$51 thousand Bachelor’s degree = \$89 thousand Step 5: Check the proposed solution in the wording of the problem. The solution checks.
10. The total price of an article purchased on a monthly deferred payment plan is described by the following formula: T is the total price, D is the down payment, p is the monthly payment, and m is the number of months one pays. Solve the formula for p. T – D = D – D + pm T – D = pm T – D = pm m m T – D = p m Example 6: Solving a Formula for One of its Variables | 697 | 2,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-23 | latest | en | 0.900371 |
http://mymathforum.com/number-theory/14142-clock-hands.html | 1,555,597,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517682.16/warc/CC-MAIN-20190418141430-20190418163430-00011.warc.gz | 120,190,275 | 9,576 | My Math Forum the clock hands
Number Theory Number Theory Math Forum
August 15th, 2010, 12:05 PM #1 Newbie Joined: Aug 2010 Posts: 3 Thanks: 0 the clock hands I don't know if it's the right place to ask this question so I'm sorry if it's not. I'm a rather layperson in mathematics but there's one puzzle I've been given by my friend recently and which i have to solve in order to borrow something from her. Can you help me with this, please? It goes like this: "When I checked my watch this morning the hour hand was exactly where the minute hand is now, and the minute hand was one minute division before where the hour hand is now. And both hands are exactly at minute divisions now. What was the time when I checked it in the morning?" I went like: x=the place of the hour hand now, y=the place of the minute hand now, likewise z=the place of the hour hand earlier, w=the place of the minute hand earlier So: (60z + w)/12=y and (60x + y)/12 - 1=w It made the final equation 720z + 60x - 143y = 12 What should I do next? I don't know how to do the equations with three unknowns (though the unknowns are all integers) Anyone can please explain how I should proceed? I would appreciate that.
August 15th, 2010, 09:34 PM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: the clock hands 7:12. At 2:36 the hour hand is 13 ticks past 12 and the minute hand is 36 ticks past twelve. In the morning the minute hand was 12 ticks past 12 and the hour hand was 36 ticks past 12.
August 16th, 2010, 01:52 AM #3 Newbie Joined: Aug 2010 Posts: 3 Thanks: 0 Re: the clock hands Thanks a lot! I got what I wanted. How did you do that? you did any calculations or you just knew it straight away? I'm just curious...
August 16th, 2010, 09:25 AM #4 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: the clock hands I wrote p = b - 1 q = a a b and 12 24 36 48 00 then the answer became apparent.
August 16th, 2010, 10:47 AM #5 Newbie Joined: Aug 2010 Posts: 3 Thanks: 0 Re: the clock hands Right...foolish me give me some of your brains Greg please...Thanks once again.
Tags clock, hands
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post Liwayway New Users 0 March 21st, 2014 10:11 AM josh2499 Algebra 2 March 28th, 2012 07:26 AM TheMytique Applied Math 2 April 18th, 2011 12:06 PM umang New Users 1 September 9th, 2010 08:02 PM Nicolaos Advanced Statistics 1 June 12th, 2009 02:52 PM
Contact - Home - Forums - Cryptocurrency Forum - Top | 772 | 2,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-18 | latest | en | 0.953468 |
https://www.woodsideacademy.co.uk/week-6-061221-101221-1/ | 1,659,996,627,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00561.warc.gz | 932,088,943 | 70,260 | Search
Close
Translate
Close
# Week 6 06.12.21-10.12.21
Literacy:
Our Story of the Week is 'Dear Santa' by Rod Campbell. Watch the video below and while watching ask your child to predict what might be hiding underneath the presents. Highlight the language - too small, too big and too scary. Can they predict what the final gift could be.
Can you write a letter to Santa? Choose some nice paper or card and tell a grown up what you would like Santa to bring you this year. Is there anything else you would like? Maybe you could have a go at writing your name at the end. You could even post your letter off to Santa when you are finished.
## Dear Santa by Rod Campbell - With fun questions!
'Dear Santa' by Rod Campbell.
Maths:
Our number song this week is 'One ,two Buckle my shoe. By singing number rhymes children will learn how to count. Have fun joining in with the rhyme.
## One Two Buckle My Shoe
This week in maths we are learning to separate a group of three or four of objects in different ways and we are beginning to to recognise that the total is still the same.
Separating objects means a variety of different ways to split and arrange a set of objects. The total does not change. Watch online, click on the video below.
## Separates a group of objects in different ways and understand the total is still the same
Challenge:
1. Gather a collection of four soft toys and two pillows or cushions.
2. Explain that it is time for the toys to go to bed.
3. Encourage the child to arrange the toys on the pillows in different ways and then count the total number of toys. This will encourage them to recognise that the total number stays the same, no matter how the toys are arranged.
How to Get Your Child Thinking:
Try asking questions, such as: How many toys are on this pillow? How many toys altogether? If we move the toys all on to this pillow, how many are there? What can you tell me about the number of toys? Have a go at using three pillows or cushions instead of two. Again, encourage the child to arrange the four toys between the three pillows and count the total. Talk about what they notice.
Phonics:
Children in nursery are beginning to learn some phase 2 sounds and are practicing blending sounds together. Watch the video below and practice saying the sounds.
Top | 516 | 2,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-33 | latest | en | 0.923569 |
https://www.physicsforums.com/threads/harmonic-oscillator.912578/ | 1,508,426,387,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00625.warc.gz | 958,367,425 | 16,931 | Harmonic Oscillator
1. Apr 25, 2017
alex91alex91alex
1. The problem statement, all variables and given/known data
I am having issues with d) and would like to know if I did the a, b, and c correctly. I have tried to look online for explanation but with no success.
A harmonic oscillator executes motion according to the equation x=(12.4cm)cos( (34.4 rad /s)t+ π/5 ) .
a) Determine the amplitude of the oscillation.
b) Determine the maximum velocity of the oscillation.
c) Determine the period of the oscillation.
d) Determine when the object is at its equilibrium position.
2. Relevant equations
max v=Aw
T=1/f
3. The attempt at a solution
a) Determine the amplitude of the oscillation.
Amplitude would just be 12.4cm, we can take it straight out from the equation.
b) Determine the maximum velocity of the oscillation.
c) Determine the period of the oscillation.
Here we know that 2π rad is one turn so 34.4 rad is 5.48 turns.
So we have 5.48 turns per second.
T=1/f=1/5.48=0.183secs
d) Determine when the object is at its equilibrium position.
I know that the object is in equilibrium when x=0, so
0=(12.4cm)cos( (34.4 rad /s)t+ π/5 )
I may be missing some algebra skills, but I even try to compute it online and it wields no solution. What am I doing wrong?
2. Apr 25, 2017
Staff: Mentor
Your work on parts (a) through (c) looks fine.
For part (d), consider what the argument of the cosine function needs to be for the cosine to be zero. (Hint: there are many such angles)
3. Apr 25, 2017
alex91alex91alex
So cos( (34.4 rad /s)t+ π/5 ) need to equal 0?
(34.4rad/s)t+π/5 has to be equal to π/2 or 3π/2?
4. Apr 25, 2017
alex91alex91alex
That is going to wield t= 0.02739s and 0.1187s, does not feel right though because a period takes 0.183secs.
5. Apr 25, 2017
Staff: Mentor
Yes.
In fact, the cosine will be zero every time its argument is equivalent to π/2 or 3π/2. You should be able to write it as a function of n, where n = 0,1,2,.... Or you can solve for the first instance (n = 0 so that the argument is π/2) and then it will happen every half period after that. | 626 | 2,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-43 | longest | en | 0.896694 |
https://us.metamath.org/mpeuni/cygabl.html | 1,709,591,255,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00286.warc.gz | 589,124,961 | 8,062 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cygabl Structured version Visualization version GIF version
Theorem cygabl 18946
Description: A cyclic group is abelian. (Contributed by Mario Carneiro, 21-Apr-2016.) (Proof shortened by AV, 20-Jan-2024.)
Assertion
Ref Expression
cygabl (𝐺 ∈ CycGrp → 𝐺 ∈ Abel)
Proof of Theorem cygabl
Dummy variables 𝑛 𝑥 𝑎 𝑏 𝑖 𝑚 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2826 . . 3 (Base‘𝐺) = (Base‘𝐺)
2 eqid 2826 . . 3 (.g𝐺) = (.g𝐺)
31, 2iscyg3 18941 . 2 (𝐺 ∈ CycGrp ↔ (𝐺 ∈ Grp ∧ ∃𝑥 ∈ (Base‘𝐺)∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)))
4 eqidd 2827 . . . 4 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → (Base‘𝐺) = (Base‘𝐺))
5 eqidd 2827 . . . 4 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → (+g𝐺) = (+g𝐺))
6 simpll 763 . . . 4 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → 𝐺 ∈ Grp)
7 oveq1 7157 . . . . . . . . . . 11 (𝑛 = 𝑖 → (𝑛(.g𝐺)𝑥) = (𝑖(.g𝐺)𝑥))
87eqeq2d 2837 . . . . . . . . . 10 (𝑛 = 𝑖 → (𝑦 = (𝑛(.g𝐺)𝑥) ↔ 𝑦 = (𝑖(.g𝐺)𝑥)))
98cbvrexvw 3456 . . . . . . . . 9 (∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥) ↔ ∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
109biimpi 217 . . . . . . . 8 (∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥) → ∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
1110ralimi 3165 . . . . . . 7 (∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥) → ∀𝑦 ∈ (Base‘𝐺)∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
1211adantl 482 . . . . . 6 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → ∀𝑦 ∈ (Base‘𝐺)∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
13123ad2ant1 1127 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → ∀𝑦 ∈ (Base‘𝐺)∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
14 simpll 763 . . . . . . . . 9 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → 𝐺 ∈ Grp)
15 simpr 485 . . . . . . . . . . 11 ((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) → 𝑥 ∈ (Base‘𝐺))
1615anim1ci 615 . . . . . . . . . 10 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → ((𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ) ∧ 𝑥 ∈ (Base‘𝐺)))
17 df-3an 1083 . . . . . . . . . 10 ((𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ ∧ 𝑥 ∈ (Base‘𝐺)) ↔ ((𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ) ∧ 𝑥 ∈ (Base‘𝐺)))
1816, 17sylibr 235 . . . . . . . . 9 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ ∧ 𝑥 ∈ (Base‘𝐺)))
19 eqid 2826 . . . . . . . . . 10 (+g𝐺) = (+g𝐺)
201, 2, 19mulgdir 18204 . . . . . . . . 9 ((𝐺 ∈ Grp ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ ∧ 𝑥 ∈ (Base‘𝐺))) → ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
2114, 18, 20syl2anc 584 . . . . . . . 8 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
2221ralrimivva 3196 . . . . . . 7 ((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) → ∀𝑚 ∈ ℤ ∀𝑛 ∈ ℤ ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
2322adantr 481 . . . . . 6 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → ∀𝑚 ∈ ℤ ∀𝑛 ∈ ℤ ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
24233ad2ant1 1127 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → ∀𝑚 ∈ ℤ ∀𝑛 ∈ ℤ ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
25 simp2 1131 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → 𝑎 ∈ (Base‘𝐺))
26 simp3 1132 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → 𝑏 ∈ (Base‘𝐺))
27 zsscn 11983 . . . . . 6 ℤ ⊆ ℂ
2827a1i 11 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → ℤ ⊆ ℂ)
2913, 24, 25, 26, 28cyccom 18291 . . . 4 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → (𝑎(+g𝐺)𝑏) = (𝑏(+g𝐺)𝑎))
304, 5, 6, 29isabld 18856 . . 3 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → 𝐺 ∈ Abel)
3130r19.29an 3293 . 2 ((𝐺 ∈ Grp ∧ ∃𝑥 ∈ (Base‘𝐺)∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → 𝐺 ∈ Abel)
323, 31sylbi 218 1 (𝐺 ∈ CycGrp → 𝐺 ∈ Abel)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 396 ∧ w3a 1081 = wceq 1530 ∈ wcel 2107 ∀wral 3143 ∃wrex 3144 ⊆ wss 3940 ‘cfv 6354 (class class class)co 7150 ℂcc 10529 + caddc 10534 ℤcz 11975 Basecbs 16478 +gcplusg 16560 Grpcgrp 18048 .gcmg 18169 Abelcabl 18843 CycGrpccyg 18932 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1904 ax-6 1963 ax-7 2008 ax-8 2109 ax-9 2117 ax-10 2138 ax-11 2153 ax-12 2169 ax-ext 2798 ax-sep 5200 ax-nul 5207 ax-pow 5263 ax-pr 5326 ax-un 7455 ax-cnex 10587 ax-resscn 10588 ax-1cn 10589 ax-icn 10590 ax-addcl 10591 ax-addrcl 10592 ax-mulcl 10593 ax-mulrcl 10594 ax-mulcom 10595 ax-addass 10596 ax-mulass 10597 ax-distr 10598 ax-i2m1 10599 ax-1ne0 10600 ax-1rid 10601 ax-rnegex 10602 ax-rrecex 10603 ax-cnre 10604 ax-pre-lttri 10605 ax-pre-lttrn 10606 ax-pre-ltadd 10607 ax-pre-mulgt0 10608 This theorem depends on definitions: df-bi 208 df-an 397 df-or 844 df-3or 1082 df-3an 1083 df-tru 1533 df-ex 1774 df-nf 1778 df-sb 2063 df-mo 2620 df-eu 2652 df-clab 2805 df-cleq 2819 df-clel 2898 df-nfc 2968 df-ne 3022 df-nel 3129 df-ral 3148 df-rex 3149 df-reu 3150 df-rmo 3151 df-rab 3152 df-v 3502 df-sbc 3777 df-csb 3888 df-dif 3943 df-un 3945 df-in 3947 df-ss 3956 df-pss 3958 df-nul 4296 df-if 4471 df-pw 4544 df-sn 4565 df-pr 4567 df-tp 4569 df-op 4571 df-uni 4838 df-iun 4919 df-br 5064 df-opab 5126 df-mpt 5144 df-tr 5170 df-id 5459 df-eprel 5464 df-po 5473 df-so 5474 df-fr 5513 df-we 5515 df-xp 5560 df-rel 5561 df-cnv 5562 df-co 5563 df-dm 5564 df-rn 5565 df-res 5566 df-ima 5567 df-pred 6147 df-ord 6193 df-on 6194 df-lim 6195 df-suc 6196 df-iota 6313 df-fun 6356 df-fn 6357 df-f 6358 df-f1 6359 df-fo 6360 df-f1o 6361 df-fv 6362 df-riota 7108 df-ov 7153 df-oprab 7154 df-mpo 7155 df-om 7574 df-1st 7685 df-2nd 7686 df-wrecs 7943 df-recs 8004 df-rdg 8042 df-er 8284 df-en 8504 df-dom 8505 df-sdom 8506 df-pnf 10671 df-mnf 10672 df-xr 10673 df-ltxr 10674 df-le 10675 df-sub 10866 df-neg 10867 df-nn 11633 df-n0 11892 df-z 11976 df-uz 12238 df-fz 12888 df-seq 13365 df-0g 16710 df-mgm 17847 df-sgrp 17896 df-mnd 17907 df-grp 18051 df-minusg 18052 df-mulg 18170 df-cmn 18844 df-abl 18845 df-cyg 18933 This theorem is referenced by: lt6abl 18951 frgpcyg 20655
Copyright terms: Public domain W3C validator | 4,230 | 6,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-10 | latest | en | 0.267355 |
http://www.docstoc.com/docs/51990803/Expectation-Maximization-Algorithm | 1,436,171,313,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375098071.98/warc/CC-MAIN-20150627031818-00136-ip-10-179-60-89.ec2.internal.warc.gz | 472,321,349 | 47,236 | # Expectation Maximization Algorithm by khn19658
VIEWS: 13 PAGES: 61
• pg 1
``` Expectation Maximization Algorithm
Rong Jin
A Mixture Model Problem
20
18
16
14
12
10
8
6
4
2
0
0 5 10 15 20 25
Apparently, the dataset consists of two modes
How can we automatically identify the two modes?
Gaussian Mixture Model (GMM)
Assume that the dataset is generated by two
mixed Gaussian distributions
Gaussian model 1: 1 1 , 1 ; p1
Gaussian model 2: 2 2 , 2 ; p2
If we know the memberships for each bin,
estimating the two Gaussian models is easy.
How to estimate the two Gaussian models
without knowing the memberships of bins?
EM Algorithm for GMM
Let memberships to be hidden variables
{x1 , x2 ,..., xn } x1 , m1 , x2 , m2 ,..., xn , mn
EM algorithm for Gaussian mixture model
Unknown memberships: x1 , m1 , x2 , m2 ,..., xn , mn
1 1 , 1 ; p1
Unknown Gaussian models:
2 2 , 2 ; p2
Learn these two sets of parameters iteratively
20
18
16
14
Random assign the
12
10
memberships to
8
6
each bin
4
2
0
1
0 5 10 15 20 25
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
20
18
16
14
Random assign the
12
10
memberships to
8
6
each bin
4
2 Estimate the means
0
1
0.9
0.8
0 5 10 15 20 25
and variance of
0.7
0.6 each Gaussian
0.5
0.4
0.3
model
0.2
0.1
0
0 5 10 15 20 25
E-step
Fixed the two Gaussian models
Estimate the posterior for each data point
p( x, m 1) p( x,1 ) p( x | 1 , 1 ) p1
p(m 1| x)
p ( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
p( x, m 2) p ( x, 2 ) p( x | 2 , 2 ) p2
p (m 2 | x)
p ( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
EM Algorithm for GMM
20
18
16
Re-estimate the
14
12 memberships for
10
8 each bin
6
4
2
0
1
0 5 10 15 20 25
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
M-Step
Fixed the memberships
Weighted by posteriors
Re-estimate the two model Gaussian
n
l p(mi 1| xi ) log p( xi ,1 ) p(mi 2 | xi ) log p( xi , 2 )
ˆ ˆ
i 1
n
p(mi 1| xi ) log p1 log p( xi | 1 , 1 ) p(mi 2 | xi ) log p2 log p( xi | 2 , 2 )
ˆ ˆ
i 1
Weighted by posteriors
i 1 p(mi 1| xi ) , i 1 p(mi 1| xi ) xi , 2 i 1 p(mi 1| xi ) xi2 2
n n n
ˆ ˆ ˆ
p1 1 1 1
i 1 p ˆ (mi 1| xi ) i 1 p
ˆ (mi 1| xi )
n n
n
i 1 p (mi 2 | xi ) i 1 p(mi 2 | xi ) xi 2 i 1 p (mi 2 | xi ) xi2
n n n
ˆ ˆ ˆ
p2 , 2 , 2 22
i 1 p(mi 2 | xi ) i 1 p(mi 2 | xi )
n n
n ˆ ˆ
EM Algorithm for GMM
20
18
16 Re-estimate the
memberships for
14
12
10
8 each bin
6
4
2
Re-estimate the
0
1
0.9
0 5 10 15 20 25 models
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
At the 5-th Iteration
20
18
16
Red Gaussian
14
12
component slowly
10
8
shifts toward the left
6
4
end of the x axis
2
0
0.9
0 5 10 15 20 25
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
At the10-th Iteration
20
18
16
14
12
Red Gaussian
10
8
component still
6
4
slowly shifts
2
0
0.9
toward the left end
0 5 10 15 20 25
0.8
of the x axis
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
At the 20-th Iteration
20
18
16
Red Gaussian
14
12
component make
10
8
more noticeable shift
toward the left end of
6
4
2
0
1
0 5 10 15 20 25
the x axis
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
At the 50-th Iteration
20
Red Gaussian
18
16
component is close
14
12
10
8
to the desirable
6
4
location
2
0
1
0 5 10 15 20 25
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
At the 100-th Iteration
20
18
16 The results are
14
12 almost identical to
10
8 the ones for the
6
4 50-th iteration
2
0
1
0 5 10 15 20 25
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 20 25
EM as A Bound Optimization
EM algorithm in fact maximizes the log-likelihood function
of training data
Likelihood for a data point x
p( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
Log-likelihood of training data
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
EM as A Bound Optimization
EM algorithm in fact maximizes the log-likelihood function
of training data
Likelihood for a data point x
p( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
Log-likelihood of training data
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
EM as A Bound Optimization
EM algorithm in fact maximizes the log-likelihood function
of training data
Likelihood for a data point x
p( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
Log-likelihood of training data
l 1 , 2 i 1 log p( xi ) i 1 log p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
Logarithm Bound Algorithm
0
l (1 ,2 )
10 , 2
0
Logarithm Bound Algorithm
Point
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 ) Touch point: Q(1 10 , 2 20 ) 0
1 , 2
0 0
Logarithm Bound Algorithm
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 ) Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2
Logarithm Bound Algorithm
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (11 , 2 ) Q(1 , 2 )
1
Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2 1 , 2
2 2
• Repeat the procedure
Logarithm Bound Algorithm
Optimal
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2 1 , 2
2 2
,... • Repeat the procedure
• Converge to the local optimal
EM as A Bound Optimization
Parameter for previous iteration: 1' , 2
'
Parameter for current iteration: 1 , 2
Compute Q(1 , 2 )
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1 log
n
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p ' p( x | ' , ' ) p ' p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
log
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p ( x | ' , ' ) p ' p ( x | ' , ' ) p ' log p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
n p ( xi | 1 , 1 ) p1 p( xi | 2 , 2 ) p2
i 1
p (1' | xi ) log p ( 2
'
| xi ) log '
p ( xi | 1' , 1' ) p1
'
p( xi | 2 , 2 ) p2
' '
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1 log
n
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p ' p( x | ' , ' ) p ' p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
log
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p ( x | ' , ' ) p ' p ( x | ' , ' ) p ' log p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
n p ( xi | 1 , 1 ) p1 p( xi | 2 , 2 ) p2
i 1
p (1' | xi ) log p ( 2
'
| xi ) log '
p ( xi | 1' , 1' ) p1
'
p( xi | 2 , 2 ) p2
' '
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
Concave property 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p ( xi | 1 , 1 ) p1
p ( xi | of
' logarithm function
' ' ' ' ' ' ' '
i 1 log
n
log( p (1 p) ) p log (1 ' p) log '
p( x | , ) p '
p ( xi | 2 , 2 ) p2
0 p , 0 ' ' i ' 2 2 2 ' ' '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p ( xi | 2 , 2 ) p2
1, ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
log
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p ( x | ' , ' ) p ' p ( x | ' , ' ) p ' log p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
n p ( xi | 1 , 1 ) p1 p( xi | 2 , 2 ) p2
i 1
p (1' | xi ) log p ( 2
'
| xi ) log '
p ( xi | 1' , 1' ) p1
'
p( xi | 2 , 2 ) p2
' '
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p( xi | 1 , 1 ) p1
' ' '
p ( xi | 1 , 1 ) p1
p( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1 log
n
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p' p( x | ' , ' ) p ' p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
' Definition of posterior
p( xi | 1 , 1 ) p1
' '
p( xi | 1 , 1 ) p1
p( xi | '1 , 1 ) p1 ' log
' ' '
n p ( xi | 1 , 1 ) p1 'p ( xi | 2 , 2 ) p2' i m 1 p
' ' ' ' ' ' '
p( x | p(1 , 1 )1| 1xi ;1' , 2 )
'
i 1 p( xi | 1 , '1 ) p1' p( xi | 2 , 2 ) p2
' ' ' '
p( xi | 2 , 2 ) p2
'
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p ' p( x | ' , ' ) p ' log
i 1 1 1 i 2 2 2 p( xi | 2 , 2 ) p2
' ' '
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
i 1
p (m1 1| xi ;1' , 2 ) log
'
p(m1 2 | xi ;1' , 2 ) log
'
'
p ( xi | 1 , 1 ) p1
' ' '
p ( xi | 2 , 2 ) p2
' '
Log-Likelihood of EM Alg.
-375
-380
-385
Loglikelhood
-390
-395
-400
-405
-410
0 10 20 30 40 50 60 70 80 90 100
Iteration
Maximize GMM Model
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
What is the global optimal solution to GMM?
n
x
i 1 i
1 x1 , 1 0, 1 , 2 1, p1 p2 0.5
n
Maximizing the objective function of GMM is ill-
posed problem
Maximize GMM Model
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
What is the global optimal solution to GMM?
n
x
i 1 i
1 x1 , 1 0, 1 , 2 1, p1 p2 0.5
n
Maximizing the objective function of GMM is ill-
posed problem
Identify Hidden Variables
For certain learning problems, identifying hidden variables is
Consider a simple translation model
For a pair of English and Chinese sentences:
e : (e1 , e2 ,..., es ) c : (c1, c2 ,..., cl )
A simple translation model is
Pr(e | c ) j 1 Pr(e j | c ) j 1
s s
t
k 1
Pr(e j | ck )
The log-likelihood of training corpus e1 , c1 ,..., en , cn
l i 1 log Pr(ei | ci ) i 1 j i1 log
n n e
ci
k 1
Pr(ei , j | ci ,k )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
EM Algorithm for A Translation Model
Introduce an alignment variable for each translation pair
e1 , c1 , a1 , e2 , c2 , a2 ,..., en , cn ,, an
EM algorithm for the translation model
E-step: compute the posterior for each alignment variable Pr(a j | e j , c j )
M-step: estimate the translation probability Pr(e|c)
|e j | |e j |
Pr(a, e j , c j ) Pr(e j ,k | c j ,a (k ) ) Pr(e j ,k | c j ,a (k ) )
k 1 k 1
Pr(a | e j , c j )
a ' Pr(a ', e j , c j ) |e j | |e j |
a ' Pr(e j ,k | c j ,a '(k ) ) t 1 Pr(e j ,s | c j ,t )
ci
k 1 k 1
EM Algorithm for A Translation Model
Introduce an alignment variable for each translation pair
e1 , c1 , a1 , e2 , c2 , a2 ,..., en , cn ,, an
EM algorithm for the translation model
E-step: compute the posterior for each alignment variable Pr(a j | e j , c j )
M-step: estimate the translation probability Pr(e|c)
|e j | |e j |
Pr(a, e j , c j ) Pr(e j ,k | c j ,a (k ) ) Pr(e j ,k | c j ,a (k ) )
k 1 k 1
Pr(a | e j , c j )
a ' Pr(a ', e j , c j ) |e j | |e j |
a ' Pr(e j ,k | c j ,a '(k ) ) t 1 Pr(e j ,s | c j ,t )
ci
k 1 k 1
We are luck here. In general, this step can be extremely
difficult and usually requires approximate approaches
Compute Pr(e|c)
First compute Pr(e | c; ei , ci )
Pr(e | c; ei , ci ) (e ei ) (c ci ) a Pr(a | ei , ci ) (a(e) c)
(e e ) (c c )
a Pr(a, ei , ci ) (a(e) c)
i i
Pr(ei , ci )
|e j |
t 1 Pr(e j , s | c j ,t )
ci
Pr(e | c)
k 1^ e j ,k e
(e ei ) (c ci ) |e j |
t 1 Pr(e j ,s | c j ,t )
ci
k 1
Pr(e | c)
(e ei ) (c ci )
t 1 Pr(e | c j ,t )
ci
Compute Pr(e|c)
First compute Pr(e | c; ei , ci )
Pr(e | c; ei , ci ) (e ei ) (c ci ) a Pr(a | ei , ci ) (a(e) c)
(e e ) (c c )
a Pr(a, ei , ci ) (a(e) c)
i i
Pr(ei , ci )
|e j |
t 1 Pr(e j , s | c j ,t )
ci
Pr(e | c)
k 1^ e j ,k e
(e ei ) (c ci ) |e j |
t 1 Pr(e j ,s | c j ,t )
ci
k 1
Pr(e | c)
(e ei ) (c ci )
t 1 Pr(e | c j ,t )
ci
Pr(e | c) i 1 Pr(e | c; ei , ci )
n
Bound Optimization for A Translation Model
θ Pr(e | c) for the current iteration
θ ' Pr'(e | c) for the previous iteration
l (θ) i 1 log Pr(ei | ci ; θ) i 1 j i1 log
n n e
ci
k 1
Pr(ei , j | ci ,k )
l (θ ') i 1 log Pr(ei | ci ; θ ') i 1 j i1
n n e
log
ci
k 1
Pr'(ei , j | ci ,k )
ci Pr(e | c )
Q(θ, θ ') l (θ) l (θ ') i 1 j i1 log k 1
n e i, j i ,k
ci Pr'(e | c )
l 1 i, j i ,l
Bound Optimization for A Translation Model
ci Pr(e | c )
Q(θ, θ ') l (θ) l (θ ') i 1 j i1 log k 1
n e i, j i ,k
ci Pr'(e | c )
l 1 i, j i ,l
ci Pr'(ei , j | ci ,k ) Pr(ei , j | ci ,k )
log
n ei
i 1 j 1
k 1 ci Pr'(e | ci ,l ) Pr'(ei , j | ci ,k )
l 1 i, j
Pr'(ei , j | ci ,k ) Pr(ei , j | ci ,k )
n ei ci
log
i 1 j 1 k 1 Pr'(ei , j | ci ,k )
ci
l 1
Pr'(ei , j | ci ,l )
Pr'(e | c)
Pr(e | c) i 1 (e ei ) (c ci )
n
ci
t 1
Pr'(e | c j ,t )
Iterative Scaling
Maximum entropy model
exp( x wy ) exp( xi wyi )
, l ( Dtrain )
N
p ( y | x ; ) log
y exp( x wy ) i 1
y exp( xi wy )
Iterative scaling
All features xi , j 0
Sum of features are constant j 1 xi, j g
d
Iterative Scaling
Compute the empirical mean for each feature of every class,
i.e., ey, j N xi, j ( y, yi ) N for every j and every class y
i 1
Start w1 ,w2 …, wc = 0
Repeat
Compute p(y|x) for each training data point (xi, yi) using w from the
previous iteration
Compute the mean of each feature of every class using the estimated
probabilities, i.e., my, j N xi, j p( y | xi ) N for every j and every y
i 1
Compute w j , y
1
g
log e j , y log m j , y for every j and every y
Update w as w j , y w j , y w j , y
Iterative Scaling
w1 , w2 ,..., wc : parameters for the current iteration
' w1 , w2 ,..., wc : parameters for the last iteration
' ' '
exp( x wy )
p ( y | x ; )
y exp( x wy )
exp( xi wyi )
l ( ) p ( y | x ; )
N N
log log
i 1 i 1
y exp( xi wy )
exp( xi w'yi )
l ( ') p( y | x; ')
N N
log log
i 1 i 1
y exp( xi w'y )
exp( x w ) exp( xi w'yi )
l ( ) l ( ')
N i yi
log '
y exp( xi wy ) y exp( xi wy )
i 1
Iterative Scaling
exp( x w ) exp( xi w'yi )
l ( ) l ( ')
N i yi
log
y exp( xi wy ) y
i 1
exp( xi w'y )
i 1 xi wyi w'yi log
N
y
exp( xi w'y ) log exp( x w )
y i y
Can we use the concave property of logarithm function?
No, we can’t because we need a lower bound
Iterative Scaling
log x x 1 log exp(x w ) exp(x w ) 1
y i y y i y
l ( ) l ( ')
i 1 xi wyi w'yi log
N
y exp( x w )
exp( xi w'y ) log y i y
x w log exp( x w ) exp( x w ) 1
N
i 1 i yi w'yi y i
'
y y i y
• Weights w y still couple with each other
• Still need further decomposition
Iterative Scaling
exp q p p exp q for i, p 0, p 1
i i i i i i i i i
exp( xi wy ) exp d
x w
j 1 i , j y , j
d
exp j 1 d
xi , j
k 1 xi,k
wy , j k 1 xi ,k
d
j 1 d
d xi , j
xi,k
d d xi , j
exp wy , j k 1 xi , k j 1
g
exp gwy , j
k 1
l ( ) l ( ')
N
i 1 xi wyi w'yi log y
exp( xi w'y ) y exp( xi wy ) 1
N
j xi, j log
xi , j
i 1
wyi , j w'yi , j exp( xi w'y ) exp( gwy , j ) 1
y y j g
Iterative Scaling
N
xi , j
Q( , ') i 1 j xi , j wyi , j w'yi , j log y exp( xi w'y ) y j exp( gwy , j ) 1
g
N
log
i 1 y
exp( xi w'y ) 1 N
i 1
y j xi, j wy, j w'y, j
y, yi
xi , j
g
exp( gwy , j )
Q( , ')
wy , j
i 1 y j xi , j y, yi xi , j exp( gwy , j ) 0
N
y j xi, j y, yi
N
i 1
wy , j log
i 1 y j xi, j
N
Wait a minute, this can not be right! What happens?
Logarithm Bound Algorithm
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 ) Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2
Iterative Scaling
Q( , ')
N
i 1 log y
exp( xi w'y ) 1 N
i 1
y j
'
xi , j wy , j wy , j y, yi
xi , j
g
exp( gwy , j )
Q( ', ')
N
i 1
log y
1 y j xi , j wy , j wy , j y, yi
exp( xi w'y )
N
i 1
' '
xi , j
g
'
exp( gwy , j )
i 1
N
log y exp( xi wy ) 1 i 1 y j
'
N xi , j
g
exp( gw'y , j )
0
Where does it go wrong?
Iterative Scaling
log x x 1 log exp(x w ) exp(x w ) 1
y i y y i y
l ( ) l ( ')
N
i 1 xi wyi w'yi log y exp( x w )
exp( xi w'y ) log y i y
i 1
N
x w
i yi w'yi log y
exp( x w ) exp( x w ) 1
i
'
y y i y
Not zero when = ’
exp( xi wy ) exp( xi wy )
log x x 1 log y 1
exp( xi wy )
' y
y exp( xi wy )
'
y
Iterative Scaling
exp( xi wy ) exp( xi wy )
log x x 1 log y 1
exp( xi wy )
' y
y exp( xi wy )
'
y
l ( ) l ( ')
exp( xi wy )
N
i 1 i
x
wyi w'yi y
y
1
exp( xi w'y )
y Definition y ' conditional
wy w of
N exp( xi w'y ) exp( xi y )
i 1 xi yi y
exponential model 1
y exp( xi w'y )
N
i 1 x
i yi
y p( y | xi ; ') exp( xi y ) 1
Iterative Scaling
exp( xi y ) exp d
x
j 1 i , j y , j
exp j 1 d
d xi , j
k 1 xi,k
y , j k 1 xi ,k
d
j 1 d
d xi , j
xi,k
d
d xi , j
exp y , j k 1 xi , k j 1
g
exp g y , j
k 1
l ( ) l ( ')
N
i 1 x
i yi y p( y | xi ; ') exp( xi y ) 1
N xi , j
i 1
xi , j yi , j y p( y | xi ; ') j exp( g y , j ) 1
j
g
xi , j
i 1 j y xi , j y , j ( y, yi ) p( y | xi ; ')
N
exp( g y , j ) 1
g
Iterative Scaling
xi , j
Q( , ') jy
N
i 1
xi , j y , j ( y, yi ) p( y | xi ; ') exp( g y , j ) 1
g
Q( , ')
y , j
i 1 xi , j y , j ( y, yi ) p( y | xi ; ') xi , j exp( g y , j ) 0
N
N
1 x ( y, yi )
i 1 i , j y , j
y , j wy , j w'y , j log
N
g
i 1
p( y | xi ; ') xi , j
Iterative Scaling
How about d1 xi, j gi constant ?
j
exp( xi y ) exp d
x
j 1 i , j y , j
exp j 1 d
d xi , j
k 1 xi,k
y , j k 1 xi ,k
d
j 1 d
d xi , j
xi,k
exp y , j k 1 xi , k j 1
d
d xi , j
gi
exp gi y , j
k 1
xi , j
Q( , ') jy
N
i 1
xi , j y , j ( y, yi ) p( y | xi ; ') exp( gi y , j ) 1
gi
Q( , ')
y , j
i 1 xi , j y , j ( y, yi ) p( y | xi ; ') xi, j exp( gi y , j ) 0
N
Is this solution unique?
Iterative Scaling
exp( xi y ) exp j 1 xi, j y, j
d
d 1
d
d xi , j
exp j 1 d y , j xi , k j 1
d
exp d y , j xi , k
1
Q( , ') i 1 j y xi , j y , j ( y, yi ) p( y | xi ; ') exp( xi , j y , j d ) 1
N
d
Q( , ')
y , j
i 1 xi , j y , j ( y, yi ) p( y | xi ; ') exp(d y , j xi , j ) 0
N
Faster Iterative Scaling
The lower bound may not be tight given all the
coupling between weights is removed
xi , j
Q( , ') jy
N
i 1
xi , j y , j ( y, yi ) p( y | xi ; ') exp( gi y , j ) 1
gi
i 1 j y q( y , j )
N
Univariate functions!
A tighter bound can be derived by not fully
decoupling the correlation between weights
xi, j g
Q( , ') y, yi xi , j y , j log p( y | x)e y , j i
j i, y i gi
y
Faster Iterative Scaling
Log-likelihood
You may feel great after the struggle of the derivation.
However, is iterative scaling a true great idea?
Given there have been so many studies in optimization, we
should try out existing methods.
Comparing Improved Iterative Scaling to
Newton’s Method
Dataset Iterations Time (s)
Dataset Instances Features
Rule 823 42.48
Rule 29,602 246
81 1.13
Lex 42,509 135,182
Try out the standard numerical 241
Lex 102.18
Summary 24,044 198,467
methods before you get excited 176 20.02
Shallow 8,625,782 264,142 | 18,549 | 37,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2015-27 | longest | en | 0.590513 |
https://www.physicsforums.com/threads/spaningset-theorem.79024/ | 1,519,545,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816178.71/warc/CC-MAIN-20180225070925-20180225090925-00612.warc.gz | 925,563,553 | 15,629 | # Spaningset theorem
1. Jun 14, 2005
### Mathman23
Spaning set theorem (new Question)
Hi
I have a couple of questions regarding an assignment which deals with the spanning set theorem.
Hope You can help
The matrix $$A = [a1 \ a2 \ a3 \ a4 \ a5] = \left[ \begin{array}{ccccc} 1 & 0 & 0 & 1 & -1 \\ 0 & 1 & 1 & 2 & 0 \\ 1 & 0 & 0 & 1 & 2 \\ -1 & 2 & 2 & 3 & 1 \end{array} \right]$$
a) First I determin the rank of A rank(A) = 3
The dimension of Null A: dim (Null A) = 2
b) Determin a basis for A's column space.
I do this using the spanning set theorem.
Since a3 = a2, a4 = 2a2 + a1 then
B(ColA) = sp{a1,a2, a5} According to the theorem.
c) Next the basis for the B(Null A).
First I row reduce A and then up and write up the set of solutions for A which results in the set B(Null A) = span{(0,-1,1,0) , (1,-2,0,1)}
Is that the correct approach ??
d) There is a vector x = a1 + a2 + a3 + a4 + a5. I'm tasked with showing that this vector belongs to Col A. Finally I'm tasked with finding the vector x with respect to the basis B.
I need some assistance is solving c) and d) therefore I hope there is somebody out there who can guide me :-)
Sincerely and Best Regards,
Fred
/Fred
Last edited: Jun 14, 2005 | 412 | 1,223 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-09 | longest | en | 0.864837 |
https://edugnosis.com/lessons/index.php?cnt=3&main=19&subw=121 | 1,631,894,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00467.warc.gz | 278,710,832 | 2,571 | # Home | Mathematics | Pre-Algebra
## Sum of Two Cubes
The other two special factoring formulas you'll need to memorize are very similar to one another; they're the formulas for factoring the sums and the differences of cubes. Here are the two formulas:
Factoring a Sum of Cubes:
a3 + b3 = (a + b)(a2 – ab + b2)
Factoring a Difference of Cubes:
a3 – b3 = (a – b)(a2 + ab + b2)
You'll learn in more advanced classes how they came up with these formulas. For now, just memorize them.
To help with the memorization, first notice that the terms in each of the two factorization formulas are exactly the same. Then notice that each formula has only one "minus" sign. The distinction between the two formulas is in the location of that one "minus" sign:
For the difference of cubes, the "minus" sign goes in the linear factor, a – b; for the sum of cubes, the "minus" sign goes in the quadratic factor:
a2– ab + b2. | 237 | 918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | longest | en | 0.910988 |
http://nrich.maths.org/public/leg.php?code=31&cl=2&cldcmpid=1181 | 1,505,874,041,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686117.24/warc/CC-MAIN-20170920014637-20170920034637-00250.warc.gz | 246,905,318 | 10,005 | # Search by Topic
#### Resources tagged with Addition & subtraction similar to Traffic Lights:
Filter by: Content type:
Stage:
Challenge level:
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Connect Three
##### Stage: 3 and 4 Challenge Level:
Can you be the first to complete a row of three?
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
##### Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Cunning Card Trick
##### Stage: 3 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Making Maths: Double-sided Magic Square
##### Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Which Symbol?
##### Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Cayley
##### Stage: 3 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Cycling Squares
##### Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### Cubes Within Cubes
##### Stage: 2 and 3 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Dodecamagic
##### Stage: 2 Challenge Level:
Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers?
### Two and Two
##### Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Rod Measures
##### Stage: 2 Challenge Level:
Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Sort Them Out (2)
##### Stage: 2 Challenge Level:
Can you each work out the number on your card? What do you notice? How could you sort the cards?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Painting Possibilities
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . .
### Wild Jack
##### Stage: 2 Challenge Level:
A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Criss Cross Quiz
##### Stage: 2 Challenge Level:
A game for 2 players. Practises subtraction or other maths operations knowledge.
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Sitting Round the Party Tables
##### Stage: 1 and 2 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions! | 2,203 | 9,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-39 | latest | en | 0.905293 |
https://howmany.cc/convert/convert-15-74-cm-to-inches/ | 1,675,221,577,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00459.warc.gz | 296,400,160 | 15,888 | # Convert 15.74 cm to inches
## How many inches is in a centimeter?
When you want to convert 15.74 cm to a number in inches, first you need to determine how many inches one centimeter is equal to.
Here’s what I can be specific: one cm equals 0.3937 inches.
## Definition:Centimeter
The centimeter unit is a base unit of length.
It equals to 10 millimeters.
This unit is used in CGS system, maps, home repaire and all areas in our life.
A single centimeter is roughly equivalent to 39.37 inches.
## Definition of Inch
The inch is a unit of length in the UK and the US customary systems of measurement. An inch is equal to 1/12 of a foot or 1/36 yard.
## How can you convert 1 cm into inches?
To convert 1 centimeter into inches, simply multiply 1cm by the conversion factor of 0.3937.
This makes it much easier to convert 15.74 cm to inches.
So, 1 cm into inches = 1 times 0.3937 = 0.3937 inches, exactly.
• What is one centimeter to inches?
• What is conversion rate cm to inches?
• How many inches is equal to 1 cm?
• What is 1 cm in inches equal?
### How do u convert 15.74 cm to inches?
From the above, you have a good grasp of cm to inches.
Here is the exact algorithm:
Value in inches = value in cm × 0.3937
So, 15.74 cm to inches = 15.74 cm × 0.3937 = 6.196838 inches
This formula can also be used to answer similar questions:
• What is the formula to convert 15.74 cm to inches?
• How do I convert cm to inches?
• How to translate cm to inches?
• How to measure cm into inches?
• What size are 15.74 cm into inches?
cm inches 15.34 cm 6.039358 inches 15.39 cm 6.059043 inches 15.44 cm 6.078728 inches 15.49 cm 6.098413 inches 15.54 cm 6.118098 inches 15.59 cm 6.137783 inches 15.64 cm 6.157468 inches 15.69 cm 6.177153 inches 15.74 cm 6.196838 inches 15.79 cm 6.216523 inches 15.84 cm 6.236208 inches 15.89 cm 6.255893 inches 15.94 cm 6.275578 inches 15.99 cm 6.295263 inches 16.04 cm 6.314948 inches 16.09 cm 6.334633 inches | 605 | 1,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-06 | longest | en | 0.815029 |
https://www.sawaal.com/quantitative-aptitude-arithmetic-ability-questions-and-answers/if-160y2-x1x-160then-160what-160is-160the-160value-160of-1601y1-160-1602y1y2-1-160_21584 | 1,656,793,288,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00292.warc.gz | 1,025,681,225 | 14,577 | 0
Q:
A) (1+x)(2-x)2x-1 B) (1-x)(2+x)x-1 C) (1+x)(2-x)1-2x D) (1+x)(1-2x)2-x
Explanation:
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
What is the average of Export (in tonnes) during the five financial years ?
A) 1279.5 B) 1552.4 C) 1025.9 D) 1279.6
Explanation:
0 26
Q:
There was 25% off on bags. A lady bought a bag and got additional 20% discount for paying in cash. She paid ₹480. What was the price tag (in ₹) on the bag?
A) 950 B) 825 C) 800 D) 750
Explanation:
1 162
Q:
What is the area of a rhombus(in cm ) whose side is 20 cm and one of the diagonals is 24 cm ?
A) 396 B) 392 C) 350 D) 384
Explanation:
0 97
Q:
The simplified value of is A circle is inscribed in a triangle ABC.It touches sides AB, BC and AC at points P, Q and R respectively. If BP = 9 cm, CQ = 10 cm and AR = 11 cm, then the perimeter (in cm)of the triangle ABC is:
A) 72.5 B) 57.5 C) 60 D) 75
Explanation:
1 106
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
What is the average of total Import and Export (in tonnes) during the five financial year ?
A) 2508.8 B) 2279.5 C) 2552.4 D) 2325.9
Explanation:
0 171
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
What is the ratio of total Imports to total Exports during 2013-2014, 2015-2016 and 2017-2018?
A) 4175 : 4011 B) 4011 : 4175 C) 3619 : 3604 D) 3604 : 3073
Answer & Explanation Answer: C) 3619 : 3604
Explanation:
0 118
Q:
In which financial year the total of the Exports and Imports is the lowest ?
A) 2013-2014 B) 2017-2018 C) 2015-2016 D) 2014-2015
Explanation:
0 156
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
A) 2016-2017 B) 2014-2015 C) 2015-2016 D) 2017-2018 | 779 | 2,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-27 | latest | en | 0.753029 |
https://notesread.com/theorem/ | 1,653,329,612,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00680.warc.gz | 483,430,035 | 23,394 | Theorem
Theorem . It is a statement that must be demonstrated within a formal system. Proving theorems is a central issue in deductive mathematics . However, how the theorems arise is the task of intuitive mathematics [1]
Summary
[ hide ]
• 1 Terminology
• 2 Definitions of theorem
• 3 Theorems in mathematical logic
• 4 Theorems in other sciences
• 5 Examples of theorems
• 6 Utterances
• 1 Usual form
• 2 Implicative form
• 3 Disjunctive form
• 7 Sources
• 8 References
Terminology
From the Latin word theorēma , a theorem is a proposition that must be proved logically from its hypothesis, resorting to axioms or other theorems previously demonstrated. This demonstration process is done using certain rules of inference.
The theorem is, therefore, a statement of importance. There are lower-level statements, such as the lemma (a statement linked to a longer theorem and guides the proof of this), the corollary ‘ (the statement that immediately follows the theorem) or the proposition (a result that is not found associated with no specific theorem); finally, scholium which is an important propositional observation.
It should be noted that, until the statement is proven, it is called either an unproven conjecture or proposition.
A theorem generally has a number of premises that must be listed or clarified beforehand. Then there is a conclusion, a mathematical statement, which is true under the given conditions. The informative content of the theorem is the relationship between the hypothesis and the thesis or conclusion.
Corollary . A logical statement that is an immediate consequence of a theorem, and can be proved using the properties of the theorem previously demonstrated.
Definitions of theorem
• A theorem is a statement that can be proved to be true within a logical framework.
• Theorem is a Propositionthat needs to be proved to be evident. For example: The sum of the angles of a triangle equals two right angles.
• Proposition that affirms a demonstrable truth.
• For every set, there is a set that does not belong to it. Proof. Let A be any set. Let D be the set of y that belong to set A, such that they fulfill the property “and does not belong to y”. …
• It is a truth not evident, but demonstrable.
Theorems in mathematical logic
A theorem requires a logical framework; This framework will consist of a set of Axioms Axiomatic System and an Inference process, which allows to derive theorems from the axioms and theorems that have been previously derived.
In Mathematical Logic and Propositional Logic , any proved statement is called a theorem. More specifically in mathematical logic a finite sequence of well-formed Formulas (well-formed logical formulas) F1, …, Fn is called proof , such that each Fi is either an axiom or a theorem that follows from two previous formulas Fj and Fk (such that j <i and k <i) using a deduction rule.
Given a proof like the one above if the final element Fn is not an axiom then it is a theorem. Summing up, the above can be formally said, a theorem is a well-formed formula, which is not an axiom, and which may be the final element of some proof, that is, a theorem is a well-formed logical formula for which there is a proof .
Theorems in other sciences
Frequently in Physics or Economics some important statements that can be deduced or justified from other basic statements or hypotheses are commonly called theorems. However, frequently the areas of knowledge where these statements appear have often not been properly formalized in the form of an axiomatic system, so the term theorem should strictly be used with caution to refer to these demonstrable or deductible statements of “more basic” assumptions.
Examples of theorems
Dual theorem. The principle of duality affirms that from any theorem or construction of projective geometry we can obtain another, known as the Dual Theorem, only the words point and line can be interchanged, also modifying the relationships between the points and the lines. So, by this principle:
• A point becomes a line.
• Aligned points become lines that pass through a point.
• Tangent lines become the point of tangency.
• A circumscribed circle becomes an inscribed circle.
The dual theorem the theorem Pascal is the theorem Brianchon , Theorem Feuerbach . The Circumference Euler or 9 points, is tangent to the inscribed circumferences and exinscrita to the triangle.
• Gauss theorem. The midpoints of the diagonals of an entire quadrilateral are in a straight line.
• Euler’s theorem. In any convex polyhedron , the number of faces plus the number of vertices is equal to the number of edges plus two. (faces + vertices = edges + 2).
One of the best known theorems is Thales’ Theorem , which indicates that, by drawing a Line parallel to any of its sides in a triangle, two similar triangles are obtained (that is, they have equal angles and their sides are proportional).
Another very popular theorem is the Pythagorean Theorem , which states that, in a right triangle, the square of the Hypotenuse (the side with the longest length and opposite to the right angle), is equal to the sum of the squares of the legs ( the two minor sides of the right triangle).
Statements
Usual way
when the hypothesis and the thesis are not properly distinguished;
as an example “an exterior angle of a triangle is equal to the sum of the nonadjacent triangle angles” [2]
Implicative form
they are stated in the form if p , then q , being p the hypothesis and q , the conclusion or thesis.
as an example: if the natural number p is a multiple of 8, then p is a multiple of 4
Disjunctive form
is stated as p or not q , p being the hypothesis and q the conclusion.
For example: the triangle ACB is a rectangle with the right angle at C or the sum of the squares of the legs is not equal to the square of the hypotenuse.
byAbdullah Sam
I’m a teacher, researcher and writer. I write about study subjects to improve the learning of college and university students. I write top Quality study notes Mostly, Tech, Games, Education, And Solutions/Tips and Tricks. I am a person who helps students to acquire knowledge, competence or virtue. | 1,460 | 6,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-21 | latest | en | 0.859529 |
http://mathhelpforum.com/advanced-statistics/99393-find-mean-variance-standard-deviation-print.html | 1,511,220,644,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806258.88/warc/CC-MAIN-20171120223020-20171121003020-00246.warc.gz | 172,762,421 | 3,030 | # Find the mean, variance, and standard deviation
Printable View
• Aug 27th 2009, 02:21 AM
extrem3
Find the mean, variance, and standard deviation
Q.7
In an assembly line production of industrial robots, gearbox assemblies can be installed in
two minute each if holes have been properly drilled in the boxes and in nice minutes if
the holes must be be redrilled. Twenty gearboxes are in stock, 2 with improperly drilled
holes. Five gearboxes must be selected from the 20 that are available for installation in the next five robots.
i) Find the probability that all 5 gearbox will fit properly. (2 marks)
ii) Find the mean, variance, and standard deviation of the time to install these 5 gearbox
I have no problem with Question 7(i), but i dont know how to solve 7(ii),
the following is my solution but somehow it seems to be incorrect, someone please help me.
http://img143.imageshack.us/img143/4549/7ii.jpg
• Aug 27th 2009, 04:56 AM
HallsofIvy
Quote:
Originally Posted by extrem3
Q.7
In an assembly line production of industrial robots, gearbox assemblies can be installed in
two minute each if holes have been properly drilled in the boxes and in nice minutes if
the holes must be be redrilled. Twenty gearboxes are in stock, 2 with improperly drilled
holes. Five gearboxes must be selected from the 20 that are available for installation in the next five robots.
i) Find the probability that all 5 gearbox will fit properly. (2 marks)
ii) Find the mean, variance, and standard deviation of the time to install these 5 gearbox
I have no problem with Question 7(i), but i dont know how to solve 7(ii),
the following is my solution but somehow it seems to be incorrect, someone please help me.
http://img143.imageshack.us/img143/4549/7ii.jpg
Well, that's not the mean of the time since you haven't use the fact that the values are 2 minutes and 9 minutes.
• Aug 27th 2009, 07:24 AM
extrem3 | 477 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-47 | longest | en | 0.91529 |
https://oeis.org/A081118 | 1,611,251,476,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00071.warc.gz | 486,063,763 | 4,548 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A081118 Triangle of first n numbers per row having exactly n 1's in binary representation. 9
1, 3, 5, 7, 11, 13, 15, 23, 27, 29, 31, 47, 55, 59, 61, 63, 95, 111, 119, 123, 125, 127, 191, 223, 239, 247, 251, 253, 255, 383, 447, 479, 495, 503, 507, 509, 511, 767, 895, 959, 991, 1007, 1015, 1019, 1021, 1023, 1535, 1791, 1919, 1983, 2015, 2031, 2039, 2043 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS T(n,n) = A036563(n+1) = 2^(n+1) - 3. Numbers of the form 2^t - 2^k - 1, 1 <= k < t. LINKS Reinhard Zumkeller, Rows n=1..150 of triangle, flattened FORMULA T(n, k) = 2^(n+1) - 2^(n-k+1) - 1, 1<=k<=n. a(n) = (2^A002260(n)-1)*2^A004736(n)-1; a(n)=(2^i-1)*2^j-1, where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Apr 04 2013 EXAMPLE Triangle begins: .......... 1 ......... ................ 1 ........ 3...5 ....... .............. 11 101 ...... 7..11..13 ..... .......... 111 1011 1101 ... 15..23..27..29 ... ...... 1111 10111 11011 11101 . 31..47..55..59..61 . . 11111 101111 110111 111011 111101. MATHEMATICA Table[2^(n+1)-2^(n-k+1)-1, {n, 10}, {k, n}]//Flatten (* Harvey P. Dale, Apr 09 2020 *) PROG (Haskell) a081118 n k = a081118_tabl !! (n-1) !! (k-1) a081118_row n = a081118_tabl !! (n-1) a081118_tabl = iterate (\row -> (map ((+ 1) . (* 2)) row) ++ [4 * (head row) + 1]) [1] a081118_list = concat a081118_tabl -- Reinhard Zumkeller, Feb 23 2012 CROSSREFS Cf. A000079, A018900, A014311, A014312, A014313, A023688, A023689, A023690, A023691, A038461, A038462, A038463. Cf. A181741 (primes), A208083, subsequence of A089633. Cf. A131094. Sequence in context: A153183 A247648 A117203 * A003255 A263321 A244365 Adjacent sequences: A081115 A081116 A081117 * A081119 A081120 A081121 KEYWORD nonn,tabl AUTHOR Reinhard Zumkeller, Mar 06 2003 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified January 21 12:45 EST 2021. Contains 340350 sequences. (Running on oeis4.) | 943 | 2,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-04 | latest | en | 0.384347 |
https://earthspot.org/geo/?search=Center_of_population | 1,660,271,172,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00764.warc.gz | 233,630,754 | 20,691 | # Center of population
For representational purposes only: The point on earth closest to everyone in the world on average was in calculated to be in Central Asia, with a mean distance of 5,000 kilometers (3,000 mi). Its antipodal point is correspondingly the farthest point from everyone on earth, and is located in the South Pacific near Easter Island, with a mean distance of 15,000 kilometers (9,300 mi). The data used by this figure is lumped at the country level, and is therefore precise only to country-scale distances, larger nations heavily skewed. Far more granular data -- kilometer level, is now available -- compares with this old "textbook" example.
In demographics, the center of population (or population center) of a region is a geographical point that describes a centerpoint of the region's population. There are several ways of defining such a "center point", leading to different geographical locations; these are often confused. [1]
## Definitions
Three commonly used (but different) center points are:
1. the mean center, also known as the centroid or center of gravity;
2. the median center, which is the intersection of the median longitude and median latitude;
3. the geometric median, also known as Weber point, Fermat–Weber point, or point of minimum aggregate travel.
A further complication is caused by the curved shape of the Earth. Different center points are obtained depending on whether the center is computed in three-dimensional space, or restricted to the curved surface, or computed using a flat map projection.
### Mean center
The mean center, or centroid, is the point on which a rigid, weightless map would balance perfectly, if the population members are represented as points of equal mass.
Mathematically, the centroid is the point to which the population has the smallest possible sum of squared distances. It is easily found by taking the arithmetic mean of each coordinate. If defined in the three-dimensional space, the centroid of points on the Earth's surface is actually inside the Earth. This point could then be projected back to the surface. Alternatively, one could define the centroid directly on a flat map projection; this is, for example, the definition that the US Census Bureau uses.
Contrary to a common misconception, the centroid does not minimize the average distance to the population. That property belongs to the geometric median.
### Median center
The median center is the intersection of two perpendicular lines, each of which divides the population into two equal halves. [2] Typically these two lines are chosen to be a parallel (a line of latitude) and a meridian (a line of longitude). In that case, this center is easily found by taking separately the medians of the population's latitude and longitude coordinates. John Tukey called this the cross median. [3]
### Geometric median
The geometric median is the point to which the population has the smallest possible sum of distances (or equivalently, the smallest average distance). Because of this property, it is also known as the point of minimum aggregate travel. Unfortunately, there is no direct closed-form expression for the geometric median; it is typically computed using iterative methods.[ citation needed]
## Determination
In practical computation, decisions are also made on the granularity (coarseness) of the population data, depending on population density patterns or other factors. For instance, the center of population of all the cities in a country may be different from the center of population of all the states (or provinces, or other subdivisions) in the same country. Different methods may yield different results.
Practical uses for finding the center of population include locating possible sites for forward capitals, such as Brasília, Nur-Sultan or Austin, and, along the same lines, to make tax collection easier. Practical selection of a new site for a capital is a complex problem that depends also on population density patterns and transportation networks.
## World
It is important to use a method that does not depend on a two-dimensional projection when dealing with the entire world. In a study from the Institut national d'études démographiques, [4] the solution methodology deals only with the globe. As a result, the answer is independent of which map projection is used or where it is centered. As described above, the exact location of the center of population will depend on both the granularity of the population data used, and the distance metric. With geodesic distances as the metric, and a granularity of 1,000 kilometers (600 mi), meaning that two population centers within 1000 km of each other are treated as part of a larger common population center of intermediate location, the world's center of population is found to lie where South Asia meets Central Asia [5] with an average distance of 5,200 kilometers (3,200 mi) to all humans. [4] The data used in the reference support this result to a precision of only a few hundred kilometers, hence the exact location is not known. Another analysis utilising city level population data found that the world's center of population is located close to Almaty, Kazakhstan. [6]
## By country
### Australia
Australia's population centroid is in central New South Wales. By 1996 it had moved only a little to the north-west since 1911. [7]
In Canada, a 1986 study placed the point of minimum aggregate travel just north of Toronto in the city of Richmond Hill, and moving westward at a rate of approximately 2 metres per day. [8]
### China
China's population centroid has wandered within southern Henan from 1952 to 2005. Incidentally, the two end point dates are remarkably close to each other. [9] China also plots its economic centroid or center of economy/GDP, which has also wandered, and is generally located at the eastern Henan borders.
### Estonia
European Countries median center of population in 2011
The center of population of Estonia was on the northwestern shore of Lake Võrtsjärv in 1913 and moved an average of 6 km northwest with every decade until the 1970s. The higher immigration rates during the late Soviet occupation to mostly Tallinn and Northeastern Estonia resulted the center of population moving faster towards north and continuing urbanization has seen it move northwest towards Tallinn since the 1990s. The center of population according to the 2011 census was in Jüri, just 6 km southeast from the border of Tallinn. [10]
### Finland
In Finland, the point of minimum aggregate travel is located in the former municipality of Hauho. [11] It is moving slightly to the south-west-west every year because people are moving out of the peripheral areas of northern and eastern Finland.
### Germany
In Germany, the centroid of the population is located in Spangenberg, Hesse, close to Kassel. [12]
### Great Britain
The centre of population in Great Britain did not move significantly in the 20th century. In 1901, it was in Rodsley, Derbyshire and in 1911 in Longford. In 1971 it was at Newhall, Swadlincote, South Derbyshire and in 2000, it was in Appleby Parva, Leicestershire. [13] [14] Using the 2011 census the population center can be calculated at Snarestone, Swadlincote. [15]
### Ireland
The center of population of the entire island of Ireland is located near Kilcock, County Kildare. This is significantly further east than the Geographical centre of Ireland, reflecting the disproportionately large cities of the east of the island ( Belfast and Dublin). [16] The center of population of the Republic of Ireland is located southwest of Edenderry, County Offaly. [17]
### Japan
The centroid of population of Japan is in Gifu Prefecture, almost directly north of Nagoya city, and has been moving east-southeast for the past few decades. [18] More recently,[ when?] the only large regions in Japan with significant population growth have been in Greater Nagoya and Greater Tokyo.
### New Zealand
New Zealand's median center of population over time
In June 2008, New Zealand's median center of population was located near Taharoa, around 100 km (65 mi) southwest of Hamilton on the North Island's west coast. [19] In 1900 it was near Nelson and has been moving steadily north (towards the Equator) ever since. [20]
### Sweden
The demographical center of Sweden (using the median center definition) is Hjortkvarn in Hallsberg Municipality, Örebro county. Between the 1989 and 2007 census the point moved a few kilometres to the south, due to a decreasing population in northern Sweden and immigration to the south. [21]
### Russia
The center of population in the Russian Federation is calculated by A. K. Gogolev to be at , 42 km (26 mi) south of Izhevsk. [22]
### United States
The mean center of the United States population (using the centroid definition) has been calculated for each U.S. Census since 1790. Over the last two centuries, it has progressed westward and, since 1930, southwesterly, reflecting population drift. For example, in 2010, the mean center was located near Plato, Missouri, in the south-central part of the state, whereas, in 1790, it was in Kent County, Maryland, 47 miles (76 km) east-northeast of the future federal capital, Washington, D.C.
## Sources
• Bellone F. and Cunningham R. (1993). "All Roads Lead to... Laxton, Digby and Longford." Statistics Canada 1991 Census Short Articles Series.
## References
1. ^ Kumler, Mark P.; Goodchild, Michael F. (1992). "The population center of Canada – Just north of Toronto?!?". In Janelle, Donald G. (ed.). Geographical snapshots of North America: commemorating the 27th Congress of the International Geographical Union and Assembly. pp. 275–279.
2. ^ "Centers of Population Computation for the United States, 1950-2010" (PDF). Washington, DC: Geography Division, U.S. Census Bureau. March 2011.
3. ^ Tukey, John (1977). . Addison-Wesley. p. 668. ISBN 9780201076165.
4. ^ a b Claude Grasland and Malika Madelin (May 2001). "The unequal distribution of population and wealth in the world" (PDF). Population et SociétéS. Institut national d'études démographiques. 368: 1–4. ISSN 0184-7783.
5. ^ exact phrase in the paper is "at the crossroads between China, India, Pakistan and Tajikistan"
6. ^ "Center of World Population". City Extremes. 2017. Retrieved 21 August 2017.
7. ^ "Figure 15: Shifts in the Australian Population Centroid*, 1911–1996". Parliament of Australia Parliamentary Library. Archived from the original on 19 August 2000. Retrieved 7 January 2009.
8. ^ "The Population Center of Canada – Just North of Toronto?!?" (PDF). Retrieved 21 April 2012.
9. ^
10. ^ Haav, Mihkel (2010) - "Eesti dünaamika 1913-1999"
11. ^ Uusirauma.fi Kaupunkilehti Uusi Rauma 03.08.2009 Päivän kysymys? Missä Rauman keskipiste? (in Finnish)
12. ^ Dradio.de Archived 24 October 2007 at the Wayback Machine (in German)
13. ^ "News Item". University of Leeds. Archived from the original on 3 October 2006. Retrieved 25 November 2007.
14. ^ "Population Centre". Appleby Magna & Appleby Parva. Archived from the original on 23 November 2007. Retrieved 25 November 2007.
15. ^ "Archived copy" (PDF). Archived from the original (PDF) on 23 July 2021. Retrieved 15 March 2020.```{{ cite web}}```: CS1 maint: archived copy as title ( link)
16. ^
17. ^
18. ^ "Our Country's Center of Population (我が国の人口重心)". Stat.go.jp. Retrieved 21 April 2012.
19. ^ "Subnational Population Estimates: At 30 June 2008 -- Commentary". Statistics New Zealand. Retrieved 11 November 2014.
20. ^
21. ^ "Sweden's demographic centre, SCB.se, 2008-03-18". Scb.se. 18 March 2008. Archived from the original on 29 March 2012. Retrieved 21 April 2012.
22. ^
23. ^ | 2,730 | 11,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-33 | longest | en | 0.935278 |
http://gmatclub.com/forum/findings-from-several-studies-on-corporate-mergers-161474.html | 1,484,760,001,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00021-ip-10-171-10-70.ec2.internal.warc.gz | 116,435,625 | 73,970 | Findings from several studies on corporate mergers : GMAT Reading Comprehension (RC)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 18 Jan 2017, 09:20
# STARTING SOON:
Open Admission Chat with MBA Experts of Personal MBA Coach - Join Chat Room to Participate.
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Findings from several studies on corporate mergers
Author Message
TAGS:
### Hide Tags
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 3631
Location: India
WE: Education (Education)
Followers: 715
Kudos [?]: 5556 [5] , given: 322
Findings from several studies on corporate mergers [#permalink]
### Show Tags
12 Oct 2013, 10:50
5
KUDOS
5
This post was
BOOKMARKED
Question 1
00:00
Question Stats:
88% (05:36) correct 13% (04:30) wrong based on 8
### HideShow timer Statistics
Question 2
00:00
Question Stats:
78% (04:36) correct 22% (00:44) wrong based on 9
### HideShow timer Statistics
Question 3
00:00
Question Stats:
67% (01:32) correct 33% (07:02) wrong based on 9
### HideShow timer Statistics
Question 4
00:00
Question Stats:
64% (04:12) correct 36% (01:01) wrong based on 11
### HideShow timer Statistics
Question 5
00:00
Question Stats:
33% (01:32) correct 67% (03:49) wrong based on 9
### HideShow timer Statistics
Question 6
00:00
Question Stats:
89% (03:55) correct 11% (00:00) wrong based on 9
### HideShow timer Statistics
Question 7
00:00
Question Stats:
60% (06:46) correct 40% (02:16) wrong based on 10
### HideShow timer Statistics
Findings from several studies on corporate mergers and acquisitions during the 1970s and 1980s raise questions about why firms initiate and consummate such transactions. One study showed, for example, that acquiring firms were on average unable to maintain acquired firms’ pre-merger levels of profitability. A second study concluded that postacquisition gains to most acquiring firms were not adequate to cover the premiums paid to obtain acquired firms. A third demonstrated that, following the announcement of a prospective merger, the stock of the prospective acquiring fi rm tends to increase in value much less than does that of the firm for which it bids. Yet mergers and acquisitions remain common, and bidders continue to assert that their objectives are economic ones.
Acquisitions may well have the desirable effect of channeling a nation’s resources efficiently from less to more efficient sectors of its economy, but the individual acquisitions executives arranging these deals must see them as advancing either their own or their companies’ private economic interests. It seems that factors having little to do with corporate economic interests explain acquisitions. These factors may include the incentive compensation of executives, lack of monitoring by boards of directors, and managerial error in estimating the value of firms targeted for acquisition. Alternatively, the acquisition acts of bidders may derive from
modeling: a manager does what other managers do.
Questions 64–70 refer to the passage above.
64. The primary purpose of the passage is to
(A) review research demonstrating the benefi ts of corporate mergers and acquisitions and examine some of the drawbacks that acquisition behavior entails
(B) contrast the effects of corporate mergers and acquisitions on acquiring firms and on firms that are acquired
(C) report findings that raise questions about a reason for corporate mergers and acquisitions and suggest possible alternative reasons
(D) explain changes in attitude on the part of acquiring firms toward corporate mergers and acquisitions
(E) account for a recent decline in the rate of corporate mergers and acquisitions
OA &OE
[Reveal] Spoiler:
64. A The research cited in the passage calls into question whether mergers and acquisitions are beneficial to firms.
B Th e passage is not concerned with comparing the relative effects of mergers and acquisitions on the acquired and acquiring firms.
C Correct. Th e passage surveys reports that question the reasons given by firms when they acquire other firms and suggests other reasons for these acquisitions.
D Th e passage does not indicate that there has been a change in the attitude of acquiring firms toward mergers and acquisitions.
E The passage does not indicate that there has been a decline in the rate of mergers and acquisitions.
65. The findings cited in the passage suggest which of the following about the outcomes of corporate mergers and acquisitions with respect to acquiring firms?
(A) They include a decrease in value of many acquiring firms’ stocks.
(B) They tend to be more benefi cial for small firms than for large firms.
(C) They do not fulfill the professed goals of most acquiring firms.
(D) They tend to be beneficial to such firms in the long term even though apparently detrimental in the short term.
(E) They discourage many such firms from attempting to make subsequent bids and acquisitions.
OA & OE
[Reveal] Spoiler:
65 A The passage suggests that the stock of acquiring firms tends to increase in value (lines 12–13), albeit less than the firm it acquires.
B The three studies cited in the passage do contrast the effects of corporate mergers on acquiring firms and on acquired firms, but the effects in question are signifi cant only insofar as they contribute to the wider investigation into why mergers take place at all.
C Correct. The passage indicates that even while acquiring firms cite economic goals, the results of the studies indicate that these goals are not being met.
D The passage makes no comparison between the long-term and short-term gains of acquiring firms.
E The passage does not indicate that firms have been affected by the results of the studies cited.
66. It can be inferred from the passage that the author would be most likely to agree with which of the following statements about corporate acquisitions?
(A) Their known benefi ts to national economies explain their appeal to individual firms during the 1970s and 1980s.
(B) Despite their adverse impact on some firms,they are the best way to channel resources from less to more productive sectors of a nation’s economy.
(C) They are as likely to occur because of poor monitoring by boards of directors as to be caused by incentive compensation for managers.
(D) They will be less prevalent in the future, since their actual effects will gain wider recognition.
(E) Factors other than economic benefit to the acquiring fi rm help to explain the frequency with which they occur.
OA & OE
[Reveal] Spoiler:
66. A The passage indicates that while mergers and acquisitions may benefit the national economy, the appeal of mergers and
acquisitions must be tied to companies’ private economic interests (lines 19–22).
B The passage makes no judgment as to the best way for firms to help channel resources from less to more effi cient economic sectors.
C The passage makes no comparison between the influence of poor monitoring by boards and that of executive incentives.
D The passage makes no prediction as to future trends in the market for mergers and acquisitions.
E Correct. The passage states that factors other than economic interests drive mergers and acquisitions.
67. The author of the passage mentions the effect of acquisitions on national economies most probably in order to
(A) provide an explanation for the mergers and acquisitions of the 1970s and 1980s overlooked by the fi ndings discussed in the passage
(B) suggest that national economic interests played an important role in the mergers and acquisitions of the 1970s and 1980s
(C) support a noneconomic explanation for the mergers and acquisitions of the 1970s and 1980s that was cited earlier in the passage
(D) cite and point out the inadequacy of one possible explanation for the prevalence of mergers and acquisitions during the 1970s and 1980s
(E) explain how modeling affected the decisions made by managers involved in mergers and acquisitions during the 1970s and 1980s
OA &OE
[Reveal] Spoiler:
67 A The passage does not mention national economies as part of an explanation for the occurrence of mergers and acquisitions.
B The passage suggests that the effect of acquisitions on national economies is not tied to any explanations for why acquisitions occur.
C The eff ect of acquisitions on national economies is not mentioned in the passage as an explanation for why acquisitions occur.
D Correct. Th e passage uses the mention of national economies as part of a larger point questioning the stated motivations behind firms’ efforts to acquire other firms.
E In the passage, modeling is unrelated to the idea that acquisitions may have a desirable effect on national economies.
68. According to the passage, during the 1970s and 1980s bidding firms differed from the firms for which they bid in that bidding firms
(A) tended to be more profi table before a merger than after a merger
(B) were more often concerned about the impact of acquisitions on national economies
(C) were run by managers whose actions were modeled on those of other managers
(D) anticipated greater economic advantages from prospective mergers
(E) experienced less of an increase in stock value when a prospective merger was announced
OA OE
[Reveal] Spoiler:
68. A The passage does not indicate whether the profitability of acquiring firms tended to be greater or less after a merger.
B The passage does not indicate that acquiring firms were concerned about the impact of their actions on national economies.
C The passage does not mention the actions of managers at firms that are being acquired.
D The passage does not discuss whether acquiring firms tended to expect greater overall economic gains than actually occurred.
E Correct. The passage indicates that the stock value of acquiring firms grew less than that of the firms they were attempting to acquire.
69. According to the passage, which of the following was true of corporate acquisitions that occurred during the
1970s and 1980s?
(A) Few of the acquisitions that f rms made were subsequently divested.
(B) Most such acquisitions produced only small increases in acquired firms’ levels of profitability.
(C) Most such acquisitions were based on an overestimation of the value of target firms.
(D) The gains realized by most acquiring firms did not equal the amounts expended in acquiring target firms.
(E) About half of such acquisitions led to long-term increases in the value of acquiring firms’ stocks.
OA &OE
[Reveal] Spoiler:
69 A The passage does not discuss post - acquisition divesting.
B The passage indicates that on average, the profitability of acquired firms fell after being acquired (lines 5–7).
C The passage does not indicate whether most acquiring firms overestimated the value of the firms they acquired.
D Correct. The passage states that for most acquiring firms the costs of buying the acquired fi rm were greater than the gains derived from acquiring it.
E The passage does not indicate what percentage of acquiring firms, if any, experienced long-term gains in their stock value.
70. The author of the passage implies that which of the following is a possible partial explanation for acquisition behavior during the 1970s and 1980s?
(A) Managers wished to imitate other managers primarily because they saw how financially beneficial other firms’ acquisitions were.
(B) Managers miscalculated the value of firms that were to be acquired.
(C) Lack of consensus within boards of directors resulted in their imposing conflicting goals on managers.
(D) Total compensation packages for managers increased during that period.
(E) The value of bidding firms’ stock increased significantly when prospective mergers were announced.
OA &OE
[Reveal] Spoiler:
70 A While the passage indicates that managers may have modeled their behavior on other managers, it does not provide a reason for why this would be so.
B Correct. The author states that one explanation for acquisition behavior may be that managers erred when they estimated the value of firms being acquired.
C The author discusses a lack of monitoring by boards of directors but makes no mention of consensus within these boards.
D The author does not discuss compensation packages for managers.
E The passage does not state how significantly the value of the bidding firm’s stock increased upon announcing a merger but only that it increased less in value than did the stock of the prospective firm being acquired.
.
[Reveal] Spoiler: Question #1 OA
[Reveal] Spoiler: Question #2 OA
[Reveal] Spoiler: Question #3 OA
[Reveal] Spoiler: Question #4 OA
[Reveal] Spoiler: Question #5 OA
[Reveal] Spoiler: Question #6 OA
[Reveal] Spoiler: Question #7 OA
_________________
“Better than a thousand days of diligent study is one day with a great teacher” – a Japanese proverb.
9884544509
Last edited by Vyshak on 03 Dec 2016, 09:50, edited 1 time in total.
If you have any questions
New!
Manager
Joined: 21 Aug 2012
Posts: 210
Concentration: General Management, Operations
Schools: HBS '19 (S)
GMAT 1: 740 Q49 V42
Followers: 0
Kudos [?]: 29 [0], given: 349
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
13 Oct 2013, 21:43
Can anyone explain 67? I did not find any of them right
Economist GMAT Tutor Instructor
Joined: 21 Aug 2013
Posts: 12
Followers: 5
Kudos [?]: 21 [1] , given: 0
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
17 Oct 2013, 23:13
1
KUDOS
Expert's post
roopika2990 wrote:
Can anyone explain 67? I did not find any of them right
D would be the correct answer choice here. First, locate the sentence you need to read to answer this question. The impact on national economies is mentioned in the first line of the second paragraph, so we need go no farther then here to find the answer. What does this sentence say about the effect on national companies? That although there may be a positive effect, this is not why such deals were arranged. This implies that some people have considered the positive effect on national economies to be an explanation of all those mergers and acquisitions. Now, notice the bare logical structure of this sentence: X may well be..., but....Y. In other words, Although X, Y. What is X - Acquisitions' impact on national economies. What is Y? The company or personal interest of executives. So what this sentence telling us? That X is not an adequate explanation of mergers and acquisitions, but Y is. This is precisely what answer choice D is telling us: cite and point out the inadequacy of one possible explanation, etc.
_________________
http://econgm.at/economistgmat
Economist GMAT Tutor
(866) 292-0660
Intern
Status: preparing for the GMAT
Joined: 16 Jul 2013
Posts: 40
Concentration: Technology, Entrepreneurship
GMAT Date: 10-15-2013
GPA: 3.53
Followers: 0
Kudos [?]: 8 [0], given: 5
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
25 Nov 2013, 06:04
daagh could you plz post the official answers?
_________________
لا الله الا الله, محمد رسول الله
You never fail until you stop trying ,,,
Manager
Joined: 19 Mar 2012
Posts: 128
Followers: 0
Kudos [?]: 12 [0], given: 270
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
28 Mar 2014, 11:26
Question 69
Also is this a 700+ RC..because I got only 3 right in like 12 mins
Or I will just accept that I have to work hard on RCs.
Intern
Joined: 19 Aug 2014
Posts: 15
Followers: 0
Kudos [?]: 2 [0], given: 122
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
01 Sep 2014, 20:45
What's considered an acceptable time frame for this passage + set of questions? Should it ideally be completed within 9 minutes? I completed it in 10:06 because I had to stop and think for a few of them. Felt like I was taking longer than I should have. Input would be great. Thanks!
Manager
Joined: 06 Jan 2014
Posts: 74
Followers: 1
Kudos [?]: 23 [1] , given: 79
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
02 Sep 2014, 05:14
1
KUDOS
Can anyone explain Q66
I am not getting how to infer the bold part in option E
Factors other than economic benefit to the acquiring firm help to explain the frequency with which they occur
_________________
______________________________
Liked the Post !!!! KUDOs Plzzzzz
Manager
Joined: 25 Apr 2014
Posts: 154
Followers: 0
Kudos [?]: 49 [2] , given: 1474
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
03 Sep 2014, 07:56
2
KUDOS
dream21 wrote:
Can anyone explain Q66
I am not getting how to infer the bold part in option E
Factors other than economic benefit to the acquiring firm help to explain the frequency with which they occur
Hey there,
If you look at the last line of para 1 it says "Yet mergers and acquisitions remain common, and bidders continue to assert that their objectives are economic ones" where "common" more or less means "frequent".
Hope this helps!
Director
Joined: 10 Mar 2013
Posts: 608
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Followers: 15
Kudos [?]: 266 [0], given: 200
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
01 Nov 2014, 05:12
maggie27 wrote:
dream21 wrote:
Can anyone explain Q66
I am not getting how to infer the bold part in option E
Factors other than economic benefit to the acquiring firm help to explain the frequency with which they occur
Hey there,
If you look at the last line of para 1 it says "Yet mergers and acquisitions remain common, and bidders continue to assert that their objectives are economic ones" where "common" more or less means "frequent".
Hope this helps!
I had also problems with this question. The statement about the frequency is wrong, because it was not mentioned in the passage, it's just an assumption. But E seems to be the only wright answer compared with other answer choices. As I see, sometimes right answers can be only derived by excluding more wrong answers.
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660
Manager
Status: GMAT Instructor
Joined: 23 Jun 2013
Posts: 189
Location: India
GRE 1: 2280 Q790 V710
GPA: 3.3
WE: Editorial and Writing (Education)
Followers: 9
Kudos [?]: 86 [0], given: 4
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
01 Nov 2014, 22:20
In the second paragraph, the author explicitly says "It seems that factors having little to do with corporate economic interests explain acquisitions."
This statement tells you that factors other than economic interests help explain acquisitions -- a logical inference is that if you know why the acquisitions occur, you can explain the frequency with which they occur with a high degree of accuracy. Hence E.
The first paragraph provides examples of firms initiating acquisitions even when they don't derive economic benefit from them, and says that acquisitions remain common. "Common" in this context seems to mean "frequent", but even this inference is not required; the inference in the above paragraph should suffice.
_________________
EnterMBA
Manager
Joined: 28 Jul 2013
Posts: 84
Followers: 0
Kudos [?]: 10 [0], given: 3
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
04 May 2015, 08:43
282552 wrote:
Question 69
Also is this a 700+ RC..because I got only 3 right in like 12 mins
Or I will just accept that I have to work hard on RCs.
D says that most such acquisitions were based on an overestimation of the value of target firms.
Actually this piece of information about overestimation of the value of target firms appears in the second para.
However, the information that we need to look for this question, is limited to the first para, since that is where it is specifically covered what the scenario was in the 70s and 80s.
VP
Joined: 08 Jun 2010
Posts: 1398
Followers: 3
Kudos [?]: 113 [0], given: 812
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
05 May 2015, 21:08
IlanaEconomistGMAT wrote:
roopika2990 wrote:
Can anyone explain 67? I did not find any of them right
D would be the correct answer choice here. First, locate the sentence you need to read to answer this question. The impact on national economies is mentioned in the first line of the second paragraph, so we need go no farther then here to find the answer. What does this sentence say about the effect on national companies? That although there may be a positive effect, this is not why such deals were arranged. This implies that some people have considered the positive effect on national economies to be an explanation of all those mergers and acquisitions. Now, notice the bare logical structure of this sentence: X may well be..., but....Y. In other words, Although X, Y. What is X - Acquisitions' impact on national economies. What is Y? The company or personal interest of executives. So what this sentence telling us? That X is not an adequate explanation of mergers and acquisitions, but Y is. This is precisely what answer choice D is telling us: cite and point out the inadequacy of one possible explanation, etc.
this passage is easy but the questions are harder. reading skill is never taught and discussed in rc forum though it is fatally important.
thank you expert, sir
Intern
Joined: 06 Nov 2014
Posts: 33
Followers: 1
Kudos [?]: 7 [0], given: 269
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
18 Aug 2015, 14:53
Hello everyone ....
I have a doubt in question 70, its pretty simple but my concern is the question's phrasing . The questions says" The author of the passage implies..... ?"
Now as far as I know if we have a imply question the answer to that question will not be explicitly mentioned in the passage and that is not case here. Answer choice B is clearly stated.Its like restating the given.
So is my thumb rule for imply questions incorrect ?? or this a strange case.
Thanks and kudos for solutions !
Jamboree GMAT Instructor
Status: GMAT Expert
Affiliations: Jamboree Education Pvt Ltd
Joined: 15 Jul 2015
Posts: 294
Location: India
Followers: 67
Kudos [?]: 246 [1] , given: 1
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
20 Aug 2015, 21:54
1
KUDOS
Expert's post
Hi Anurag,
The word IMPLIES (in this question) - is towards the word "Partial Explanation" as the answer choice B (explicitly mentioned in the passage) is only one of the three factors mentioned in the second paragraph.
Since, the choice only referes to one factor that is why "implies that ..possible partial explanation".
Hope it helps!
_________________
Aryama Dutta Saikia
Jamboree Education Pvt. Ltd.
GMAT Club Legend
Joined: 01 Oct 2013
Posts: 10531
Followers: 918
Kudos [?]: 203 [0], given: 0
Re: Findings from several studies on corporate mergers [#permalink]
### Show Tags
21 Oct 2016, 00:27
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: Findings from several studies on corporate mergers [#permalink] 21 Oct 2016, 00:27
Similar topics Replies Last post
Similar
Topics:
Findings from several studies on corporate mergers and 6 01 May 2013, 19:26
4 In corporate purchasing, competitive scrutiny is typically 4 30 Sep 2012, 01:19
33 The modern multinational corporation is described as having 23 01 Apr 2012, 18:50
8 As the economic role of multinational, global corporations 16 07 Nov 2010, 18:57
multinational corporation 1 29 Jul 2009, 17:45
Display posts from previous: Sort by | 5,838 | 24,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-04 | latest | en | 0.846001 |
https://centumwiki.in/index.php?title=Real_Numbers&diff=43&oldid=42 | 1,638,822,813,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00257.warc.gz | 233,633,237 | 7,572 | # Difference between revisions of "Real Numbers"
Concepts Lectures on Real Numbers and Divisibility
## Concept Lectures
Session # 1 What are Real Numbers?
Session # 2 What is meant by divisibility?
Session # 3 Properties of divisibility : Part 1
Session # 4 Properties of divisibility : Part 2
Session # 5 Euclid's Division Lemma : Part 1
Session # 6 Euclid's Division Lemma : Part 2
Session # 7 Euclid's Division Lemma : Part 3
Session # 8 Proof of the Euclid's Division Lemma
Session # 9 Euclid's Division Lemma - Applications
## Problems
Problem Solving Session # 1 Show that n^2-1 is divisible by 8 if n is an odd positive integer.
Problem Solving Session # 2 Show that the square of any positive integer is of the form of 3m or 3m+1 for some integer m.
Problem Solving Session # 3 Prove that one of every three consecutive positive integers is divisible by 3
## Concept Lectures
Session # 10 Euclid's Division Algorithm
Session # 11 Euclid's Division Algorithm: What is GCD?
## Problems
Problem Solving Session # 4 Find the GCD/HCF of 237 and 81 and express it as a linear combination of 237 and 81
Problem Solving Session # 5 Find the GCD/HCF of 72 and 56 and express it as a linear combination of 72 and 56. Also show that the linear combination is not unique.
Problems Solving Session # 6 Word Problem
Problems Solving Session # 7 Word Problem
## Concept Lectures
Session # 16 Fundamental Theorem of Arithmetic
## Problems
Problems Solving Session # 8 Express 168 and 234 as a product of prime factors
Problems Solving Session # 9 Prove that there is no natural number n for which 4^n ends with digit zero
Problem Solving Session # 10 Prove that there are infinitely many prime numbers
Problem Solving Session # 11Prove that a positive integer n is a prime number, if no prime less than or equal to square root of n, divides n | 480 | 1,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-49 | latest | en | 0.787967 |
https://es.mathworks.com/matlabcentral/answers/1815800-how-to-quickly-identify-a-list-of-nearby-stations | 1,670,164,413,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00174.warc.gz | 274,415,125 | 38,868 | # How to quickly identify a list of nearby stations?
3 views (last 30 days)
Leon on 2 Oct 2022
Answered: Walter Roberson on 2 Oct 2022
I have a total of 1 million stations numbered from 1 to 1 million. For each of them, how do I quickly identify the rest of the stations (Station numbers) that are within 2 km of distance? I wrote the below program, but it was extremely slow. Any recommendations would be greatly appreciated. Many thanks!
%% calculate the distance between the stations:
m2 = numel(Sta); % The list of profile numbers
B0 = zeros(m2*m2, 6);
n0 = 0;
for i=1:m2
for j=1:m2
if Sta{i}.lon <= 180
Lon1 = Sta{i}.lon;
else
Lon1 = Sta{i}.lon-360;
end
if Sta{j}.lon <= 180
Lon2 = Sta{j}.lon;
else
Lon2 = Sta{j}.lon-360;
end
n0 = n0+1;
B0(n0,:) = [i, j, Sta{i}.lat, Lon1, Sta{j}.lat, Lon2];
end
end
Dis = distance(B0(:,3), B0(:,4), B0(:,5), B0(:,6), 6371); % The radius of the earth is 6371km.
%units: km.
D = []; % cell array storing the list of stations that are close to this profile:
for i=1:m2
Ind1 = B0(:,1)==i; % Index of the distance values for Station i.
B = [B0(Ind1,2), Dis(Ind1)]; % Two columns: List of profiles, Distance.
Ind = B(:,2) <= 2; % distance is less than 2km.
if sum(Ind) >= 2 % at least two profiles identified, one of them is itself.
C0 = setdiff( B(Ind,1), i ); % list of profiles that are close to this station (excluding itself)
D{i} = C0;
else
D{i} = [];
end
end
Torsten on 2 Oct 2022
Edited: Torsten on 2 Oct 2022
And for your double loop, you should replace
for i=1:m2
for j=1:m2
by
for i=1:m2
for j=i+1:m2
to save computation time.
Image Analyst on 2 Oct 2022
You accepted this answer but we don't know how you solved the problem of needing 8000 GB of memory to hold your B0 array for your trillion distances. How did you do it?
Image Analyst on 2 Oct 2022
Well let's see why it's slow. You have a million elements and you create an array that is a million by a million by 8 bytes. So that is 8000 gigabytes! Now maybe you have an 8 terabyte drive in your computer for swap space, but it will still be slow. I don't even know if MATLAB can handle an array that big even if you had infinite memory.
The second reason it is slow is the use of cell arrays. Cell arrays are very slow and memory inefficient. Put all your locations into standard double arrays, not cell arrays.
Now I don't know what this distance function you're using is (there are several) but it might be slow because it is general purpose with lots of overhead or maybe it takes the square root or even some more complicated formula. That could be avoided by using the simple Euclidean distance formula but not taking the square root. If you want the distance to be less than 2 km, that is the same as wanting the squared distance to be less than 4. Hence just square the distances and compare to 4 but don't take the square root. That is just an extra, unnecessary operation that will slow it down.
So why do I think the square distance will probably be better? Well if you're trying to find stations that are within 2 km of each other you're probably not considering stations placed all over the globe. You're probably not worried about whether a station in Beijing is within 2 km of one in Miami. You wouldn't need that kind of precision. You're probably considering a much smaller area, like just stations in Belgium or some country or state. Though why you'd have a million stations in a small area, I don't know. What is a station anyway? A radio station? A train station? An antenna location? So if you're using a small area of the globe you can convert the lats and lons to linear (x,y) locations. But you'll say "oh but it won't be accurate" and I'll say "who cares?" Certainly not us/you. It will be accurate enough to find stations within 2 km since there is no earth curvature effect over such a small distance, and if it's inaccurate for stations separated by 500 km, who cares? You don't want to identify a pair of stations like that (very distant) anyway so who cares if it says the distance is 480 km instead of 500 km? It doesn't matter. In the end all you care about are stations where the squared distance is less than 4, not the absolutely precise distance down to the nearest millimeter.
So just have a 2-D array where you compute the squared distance of every station to every other station using the Pythagorean theorem and your (x,y) coordinates (not cell arrays). Try meshgrid. But you still have the problem of having a trillion distances to compute. You're just going to somehow reduce that down to something more manageable - a trillion distances is not realistic. So now once you have the square distances in your array, finding the ones closer than 4 is trivial
mask = (squaredDistances > 0) & (squaredDistances <= 4);
% Find station indexes
##### 2 CommentsShowHide 1 older comment
Leon on 2 Oct 2022
Wow, thank you so much for the very insightful comments! I find it very helpful.
Walter Roberson on 2 Oct 2022
First, divide the stations into subsets:
• stations close enough to the North Pole that the smaller distance between degrees of longitude could start substantially affecting the calculation of which stations are close to each other. For the moment, call this NorthStations
• A subset of NorthStations that is potentially within 2 km of the sourthern boundary of the NorthStations. Call this NorthBorderStations for the moment
• Likewise, create SouthStations and SouthBorderStations
• Take the remaining stations that are not in NorthStations or in SouthStations, and add copies of NorthBorderStations and SouthBorderStations to that. Call this OtherStations at the moment
• Take a subset of OtherStations that are potentially within 2km of -180 degrees (remember that OtherStations is clipped at 2km North/South of the boundary you decided earlier so the curve at the North/South boundary will be limited.) Add 360 to those longitudes and add that subset back into OtherStations
Now NorthStations and SouthStations are substantially reduced subsets of the original data. You can use the Haversine distance formula within each subset, perhaps using pdist(). This will compare every point in each of the subsets to every other point within the subsets, but the subsets should only be including the locations where the curve introduces notable risk of unexpected connections. The distances calculated within each of the subsets will be exact on the first pass for each of the two subsets.
Now take OtherStations and use knnsearch and Euclidean distance, with the radius set to the largest euclidean distance that maps to 2km right at the border you used for dividing into the North / South subsets. For example if you choose to put the boundary at acosd(1/3) (70.5 degrees) then use a radius corresponding to 1/(1/3)*2 = 6 km. knnsearch with these settings with create a quadtree and search it for nearby points. This will, if I recall, take roughly N * log(N) time where N is the number of points in the subset.
The list of neighbours you get will be an overestimate -- but now you can take the points and their neighbours and do pairwise havershine formulas to get exact distances.
The higher/lower you put the latitude boundaries the more you reduce the size of the NorthStations and SouthStations subsets that have to be compared pairwise, but the more you overpopulate the search in OtherStations (because the more extreme the boundary, the larger you need to make the euclidean search radius so as to be able to find the matches right at the boundary.
I guess perhaps this search strategy could end up with some duplicates, if you happen to have pairs that lie exactly within the NorthBorderStations or SouthBorderStations, or entirely within the band that is duplicated at -180, so be sure to do a final filtering pass.
Walter Roberson on 2 Oct 2022
If you have the Mapping Toolbox then use distance
Leon on 2 Oct 2022
Edited: Leon on 2 Oct 2022
The stations can be anywhere in the global ocean. Some of them will be close to the poles.
### Categories
Find more on Text Data Preparation in Help Center and File Exchange
R2022a
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 2,023 | 8,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-49 | latest | en | 0.887496 |
https://www.ques10.com/p/13168/an-lti-system-is-characterized-by-the-system-fun-1/? | 1,632,815,340,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00243.warc.gz | 954,870,229 | 7,049 | 0
1.8kviews
An LTI system is characterized by the system function:$H(z) = \dfrac {z}{(z-\frac 14)(z+\frac14)(z-\frac12)}$
write down possible ROCs for different possible ROCs. Determine causality and stability and impulse response of the system.
Mumbai University > EXTC > Sem 4 > Signals and Systems
Marks : 10
Year : DEC 2014
0
9views
$$H(z) = \dfrac z {(z-\frac14)(z+\frac14)(z-\frac12)}$$
Poles at: $z=1/4, -1/4, ½$
Pole zero plot:
Since exterior part of circle (r=1/2) lies within the unit circle. Hence the given system is causal. Also all the pole of the system lies inside the unit circle, thus having bounded output for every bounded input. Hence is a stable system
Impulse response:
$H(z) = \dfrac z {(z-\frac14)(z+\frac14)(z-\frac12)} \\ \dfrac {H(z)}z = \dfrac 1 {(z-\frac14)(z+\frac14)(z-\frac12)}$
By using partial fraction,
$\dfrac {H(z)}z = \dfrac A{(z-\frac 14)}+ \dfrac B{(z+\frac14)}+ \dfrac C{(z-\frac12)}$
Put $z=1/4$ we get, $A=-8$
Put $z=-1/4$ we get, $B=2/3$
And Put $z=1/2$ we get, $C=4/3$
$\dfrac {H(z)}z=\dfrac {-8}{(z-\frac14)} + \dfrac {2/3}{(z+\frac14)} + \dfrac {4/3}{(z-\frac12)} \\ H (z) = \dfrac {-8z}{(z-\frac14)} + \dfrac {2z/3}{(z+\frac14 )} + \dfrac {4z/3}{(z-\frac12)}$
By Inverse z-transform,
$h [n] = -8(\dfrac14)^n u(n)+\dfrac23 (\dfrac {-1}{4})^n u(n)+\dfrac43 (\dfrac12)^n u(n)$ | 524 | 1,341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-39 | latest | en | 0.604219 |
https://calculator.academy/full-load-current-calculator/ | 1,652,864,449,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00785.warc.gz | 212,822,782 | 62,036 | Enter the full load power (watts) and the full load voltage (volts) into the calculator to determine the Full Load Current.
## Full Load Current Formula
The following formula is used to calculate the Full Load Current.
If = Pf/(1.732*Vf)
• Where If is the Full Load Current (amps)
• Pf is the full load power (watts)
• Vf is the full load voltage (volts)
## How to Calculate Full Load Current?
The following two example problems outline how to calculate the Full Load Current.
Example Problem #1:
1. First, determine the full load power (watts). In this example, the full load power (watts) is measured to be 67.
2. Next, determine the full load voltage (volts). For this problem, the full load voltage (volts) is calculated to be 8.
3. Finally, calculate the Full Load Current using the formula above:
If = Pf/(1.732*Vf)
Inserting the values from above and solving the equation with the imputed values gives:
If = 67/(1.732*8) = 4.83 (amps)
Example Problem #2:
Using the same process as example problem 1, we first define the variables outlined by the formula. In this case, the values are:
full load power (watts) = 90
full load voltage (volts) = 4
Entering these values into the formula above gives :
If = 90/(1.732*4) = 12.99 (amps) | 327 | 1,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-21 | longest | en | 0.787966 |
https://drugtruthaustralia.org/and-pdf/1801-scatter-plots-and-lines-of-best-fit-worksheet-pdf-166-204.php | 1,653,195,510,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00015.warc.gz | 259,839,185 | 8,917 | Wednesday, May 26, 2021 7:23:55 PM
# Scatter Plots And Lines Of Best Fit Worksheet Pdf
File Name: scatter plots and lines of best fit worksheet .zip
Size: 19908Kb
Published: 26.05.2021
Scatter plot line of best fit worksheet.
## Sample Worksheets
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? Grade Level. Resource Type. Log In Join Us.
Line Of Best Fit - Displaying top 8 worksheets found for this concept.. There are 9 questions asking for the Slope-Intercept Form Equation of the trend line line of best fit given the scatter plot and 12 questions asking students to make a 9. Big Idea The emphasis in this lesson is to take students a little beyond the basics of Scatter Plots to explain the correlation coefficient r and the coefficient of determination r squared. Answer the questions: 80 72 56 48 40 32 16 Earnings at Work Hours Worked 1. Line of best fit: smoking in
When we have two separate data sets we can see if they have a relationship by plotting their points on in this manner. When we plot these points on an XY graph, we can see if a pattern forms. If a pattern forms, a relationship exists. We can examine this relationship using a Line of Best Fit Trend line. To create a Line of Best Fit we draw a line so that we are as close as possible to all the points.
## Print Scatter Plots and Line of Best Fit Worksheets
By: Published on: Dec 15, Categories: Uncategorized 0 comments. Approximately half of the data points should be below the line and half of the points above the line. Do not use the day on the scatter plot identify the data sets as having a positive a negative or no correlation. The teacher records the number of hours each student studied and the marks scored by the respective student on the test. The scatter plots and lines of best fit worksheet can be used to help those who have questions about scatter plots and lines of best fit.
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? Grade Level. Resource Type. Log In Join Us. View Wish List View Cart. Results for scatter plots and line of best fit.
Skip to main content. Search form Search. Lines of best fit linear regression algebra 1 homework answers. Lines of best fit linear regression algebra 1 homework answers lines of best fit linear regression algebra 1 homework answers 6 negative correlation. I: Residuals 9.
## 31 Scatter Plot And Lines Of Best Fit Worksheet
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? Grade Level. Resource Type. Log In Join Us.
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? Grade Level. Resource Type.
A Scatter XY Plot has points that show the relationship between two sets of data. The data is plotted on the graph as " Cartesian x,y Coordinates ". The local ice cream shop keeps track of how much ice cream they sell versus the noon temperature on that day.
Draw a scatter plot and determine what relationship if any exists in the data.
Или мы придем к соглашению. - Какому соглашению? - Немец слышал рассказы о коррупции в испанской полиции. - У вас есть кое-что, что мне очень нужно, - сказал Беккер.
Он не мог поверить в свою необыкновенную удачу. Он снова говорил с этим американцем, и если все прошло, как было задумано, то Танкадо сейчас уже нет в живых, а ключ, который он носил с собой, изъят. В том, что он, Нуматака, в конце концов решил приобрести ключ Энсея Танкадо, крылась определенная ирония.
Пора отсюда сматываться. - Куда ты девал мои бутылки? - угрожающе зарычал парень. В его ноздрях торчала английская булавка. Беккер показал на бутылки, которые смахнул на пол. - Они же пустые.
Стратмор знает, что я это видел! - Хейл сплюнул. - Он и меня убьет. Если бы Сьюзан не была парализована страхом, она бы расхохоталась ему в лицо.
Он решил было обратиться в полицию - может быть, у них есть данные о рыжеволосых проститутках, - но Стратмор на этот счет выразился недвусмысленно: Вы должны оставаться невидимым. Никто не должен знать о существовании кольца. Может быть, стоит побродить по Триане, кварталу развлечений, и поискать там эту рыжую девицу. Или же обойти все рестораны - вдруг этот тучный немец окажется .
Фонтейн вздохнул и обхватил голову руками. Взгляд его черных глаз стал тяжелым и неподвижным. Возвращение домой оказалось долгим и слишком утомительным. Последний месяц был для Лиланда Фонтейна временем больших ожиданий: в агентстве происходило нечто такое, что могло изменить ход истории, и, как это ни странно директор Фонтейн узнал об этом лишь случайно.
Ошеломленный потерей жены и появлением на свет неполноценного, по словам медсестер, ребенка, которому скорее всего не удастся пережить ночь, он исчез из больницы и больше не вернулся. Энсея Танкадо отдали в приемную семью. Каждую ночь юный Танкадо смотрел на свои скрюченные пальцы, вцепившиеся в куклу Дарума note 1и клялся, что отомстит - отомстит стране, которая лишила его матери, а отца заставила бросить его на произвол судьбы. Не знал он только одного - что в его планы вмешается судьба.
Так долго. Сьюзан нахмурилась, почувствовав себя слегка оскорбленной. Ее основная работа в последние три года заключалась в тонкой настройке самого секретного компьютера в мире: большая часть программ, обеспечивавших феноменальное быстродействие ТРАНСТЕКСТА, была ее творением. Шифр в миллион бит едва ли можно было назвать реалистичным сценарием.
Сьюзан отвернулась. - Не имеет значения. Кровь не .
Она была абсолютно уверена, что не вводила такой команды - во всяком случае, намеренно. Подумала, что, может быть, спутала последовательность нажатия клавиш. Немыслимо, - подумала. Согласно информации, появившейся в окне, команда была подана менее двадцати минут .
- Я ничего не знаю. Беккер зашагал по комнате. - На руке умершего было золотое кольцо.
И мы, те, кто близко к сердцу принимает интересы страны, оказались вынужденными бороться за наше право служить своей стране. Мы больше не миротворцы. Мы слухачи, стукачи, нарушители прав человека. - Стратмор шумно вздохнул. - Увы, в мире полно наивных людей, которые не могут представить себе ужасы, которые нас ждут, если мы будем сидеть сложа руки.
Кольцо? - Он вдруг забеспокоился. Вгляделся в полоску на пальце и пристыжено покраснел. - О Боже, - хмыкнул он, - значит, эта история подтверждается. Беккеру даже сделалось дурно. - Прошу прощения.
Похоже, он снискал благословение - шичигосан. Скоро Нуматек станет единственным обладателем единственного экземпляра Цифровой крепости. Другого нет и не. Двадцать миллионов долларов - это очень большие деньги, но если принять во внимание, за что они будут заплачены, то это сущие гроши. ГЛАВА 19 - А вдруг кто-то еще хочет заполучить это кольцо? - спросила Сьюзан, внезапно заволновавшись.
Все вокруг светилось ярко-красными огнями. | 2,029 | 7,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-21 | latest | en | 0.901886 |
null | null | null | null | null | null | Prima's Eliot and Jennifer Jackson-King: Chef Chat, part 1
Categories: Q&A
Michelle Bruch
The Kings work long hours to farm food for the restaurant each day.
There was a time when Prima's husband-and-wife team lived down the street from their neighborhood Italian restaurant. Now they're living on a farm outside the Cities, and the restaurant is reaping the benefits.
Chef Eliot King and his wife, Jennifer, are farming vegetables in the morning that make it onto the restaurant menu each night.
Read on to hear why the couple keeps peacocks at the farm, how the honey tastes out of the beehives, and what it takes to run a farm and restaurant simultaneously.
This is the first in a three-part series.
Prima 9.jpg
Michelle Bruch
The Prima kitchen uses the farm's honey in the "Almond-Crusted Ricotta Dolce."
Now that you're farming in addition to cooking, how long is your average workday?
Eliot: Right now my typical day starts at about 6 in the morning. I'm out with the lettuces: cutting, reseeding, and watering. Then I come into work, and I work all day. Today I won't get home until 9 or 10 o'clock at night. During the week I try to get off at 3 or 4 o'clock, and if it's really hot then I need to re-water and check on the lettuces.
Typically on the farming end of it I spend about 25 to 30 hours a week. It makes for kind of a long week.
How did you decide to take on all these extra farming chores?
Eliot: We have always farmed some tomatoes and zucchinis and cucumbers and other types of vegetables. Last summer, I was saying to myself, 'Well, I have a big field that we don't really do anything with except mow all summer. How can we make this useful and expand what we're growing?' I just haven't been thrilled with the lettuces and the greens that I've been getting out of California, especially the last two years. They just haven't been as nice as they used to be. I'm not sure exactly why.
It's kind of a test year to see how it's all going to go, and see whether we're going to expand it next year. It's been a hard year with the weather--really hot and really hard to grow lettuce, so it's been a learning experience.
How long have you lived on the farm in Waconia?
Eliot: Ten years. We live on a 20-acre farm and have horses and chickens and other types of creatures running around.
Where did you live before?
Jennifer: We lived right down the street on Lyndale. I'm a native Minnesotan and my husband's a New Yorker. We both moved here from San Francisco because we wanted to buy a house, and that was...
Eliot: 16 years ago.
Jennifer: We were going to have a child, and I always wanted her to have a somewhat rural upbringing, having had that myself. We didn't ever move out to the farm with the intention of raising food for our restaurant. It just kind of evolved.
You keep horses and chickens?
Jennifer: Horses and chickens and peacocks, and we have 20 beehives to make our own wildflower honey. You feel like you have some ownership in the product, and certainly Eliot does. From planting, nurturing, harvesting, and cooking, he has total control over when it's picked, if the flavor's right, and if the texture's right. It's pretty cool.
Why do you keep peacocks?
Jennifer: They add personality, and ironically, they're like the watchdogs of the farm. They're very astute to strange vehicles or strange creatures, and they make these horrible howling, screeching sounds when something not normal approaches. So they're kind of our bird-dog. I raised the female one myself. I took the eggs and put them in the incubator and turned them, and hatched them, and raised them in my kitchen in a little box with a light bulb.
Thumbnail image for Thumbnail image for Prima.3.jpg
Michelle Bruch
The farm's tomatoes and mixed greens show up in nightly specials at Prima.
Tell us about your beehives.
Jennifer: We have a beekeeper that we partner with, and he also does all the pollination at the Minnetonka Apple Orchard. I think he has 600 or 800 hives around the metro, and 20 at our farm. Depending on where the bees live, the honey tastes different. Ours is wildflower honey, and it's delicate and soft and lovely.
How long have you used your own produce in the restaurant?
Eliot: We've been using tomatoes and some vegetables for about five years.
Jennifer: We're realistic about what we can grow and use. I would not think that we could ever grow enough on our small family farm to supply what we need at this restaurant in its entirety. It's more of a specialized thing for us.
Eliot: It's also not what I do. I'm not so much the farmer as I am the chef. It's not like I want to spend all my time farming.
Jennifer: And we don't have outside labor at the farm either. It's all on us. It's been a rough year for tomatoes. They like dry heat. First it was cold for planting time, and then we had wet, damp weather, and then it became so humid and hot. Last year we did 400 pounds of heirloom tomatoes. I don't think we'll meet that this year--even with double the amount of plants that we're using.
What are the farm-to-table specials on the menu tonight?
Eliot: Tonight I'm doing a salad with our greens, some baby carrots from our farm, and radishes from our farm. That's with a buttermilk poppyseed dressing. Tonight I'm also doing a sautéed walleye, and I've cut some little microgreens from some mustard and kale and red amaranth. I'm using our tomatoes and corn from our neighbor, topped in a little oil with lemon verbena that also comes from our farm.
Is it fun to create specials based on what's in season at home?
Eliot: Some days I'm just waiting for stuff to grow--I'm like, 'Come on, you've got to be ready by the weekend!' This morning I went out to check on the lettuces before I came here to work, and I have a bunch of this braising mix, which has kale and mizuna and tatsoi and a bunch of little mustard greens. I've been waiting for that to grow, and [I decided] I can't wait. I'm just going to cut it for microgreens with my red amaranth and mix that together and use it. I kind of figure it out as it goes.
Location Info
5325 Lyndale Ave. S., Minneapolis, MN
Category: Restaurant
Sponsor Content
My Voice Nation Help
Greg Hoyt
Greg Hoyt
Jennifer and Eliot are wonderful people and very good operators. They treat their guests with sincerity and their employees as family. Thank you, Michelle, for doing a piece on them and Prima. They have a true jewel on Lyndale.
Now Trending
From the Vault | null | null | null | null | null | null | null | null | null |
http://www.mathcaptain.com/geometry/parabola-equation.html | 1,540,263,739,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00055.warc.gz | 495,396,764 | 10,672 | A quadratic equation always represents a parabola.Every quadratic form of equation when graphically represented, forms a parabola. When a ball is thrown or kicked, It follows a parabola.
The equation of parabola is can be written as
Where,
a, b and c are constants.
(x, y) is an arbitrary point lying on the parabola.
a $\neq$ 0. If a equals zero, then equation will be no more a quadratic equation. Hence it will not represent a parabola.
## Standard Form of a Parabola
The equation of parabola is defined above, but more generally and specifically, the standard and general form of a parabola is given below.
Where,
A, B, C, D, E and F are constants.
x and y are coordinates of an arbitrary point (x, y) lying on the parabola.
## Vertex Form of a Parabola
Sometimes, the equation of parabola is given in vertex form. The general and standard form of parabola is deducible to the vertex form of parabola. The equations of parabola in vertex form are:
Where,
(h, k) is vertex of parabola.
(x, y) is any arbitrary point on parabola.
If vertex of parabola is origin (0, 0), then h = 0 and k = 0. So, following are the parabolic equations in this case are
## Equation of Parabola
Equation of parabola can be in any of the given forms:
• Standard Form
• Vertex Form
Standard form is referred as quadratic form. It can be in below two formats:
• y = ax2 + bx + c
• Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
Vertex form can be as follows:
Vertex at origin:
• y2 = 4ax
• y2 = -4ax
• x2 = 4ay
• x2 = -4ay
Vertex at (h, k):
• (y - k)2 = 4a (x - h)
• (y - k)2 = - 4a (x - h)
• (x - h)2 = 4a (y - k)
• (x - h)2 = - 4a (y - k)
## Solving Parabola Equations
Here, let us see how to solve parabolic equations.
While solving parabola equation, following points should be kept in mind:
• Deduce given equation in standard parabolic form.
• Determine "a" by comparing it with standard form.
• Now, determine the coordinates of focus, vertex, directrix etc. according to given format.
Let us consider few examples:
### Solved Example
Question: Consider a parabola: y2 - 12x = 0. Determine its vertex and focus.
Solution:
y2 = 12x
Comparing it with standard parabola equation, y2 = 4ax:
Vertex = (0, 0)
4a = 12
a = 3
Focus = (a, 0) = (3, 0)
Example 2: Deduce the quadratic equation y2 - 4x - 3 = 0 in vertex form of parabola and find its vertex, focus and equation of directrix.
Solution: y2 - 4x - 3 = 0
y2 = 4x + 3
y2 = 4(x + $\frac{3}{4}$)
Comparing it with standard parabola equation, (y - k)2 = 4a(x - h):
Vertex = (-$\frac{3}{4}$, 0)
4a = 4
a = 1
Focus = (h + a, 0)
= (-$\frac{3}{4}$ + 1, 0)
= ($\frac{1}{4}$, 0)
Directrix:
x = -$\frac{3}{4}$ - 1
x = -$\frac{7}{4}$
## Graphing a Parabola Equation
Following steps should be followed while graphing a parabolic equation:
If equation is in quadratic form:
1. Find a, b and c by comparing it with standard quadratic form y = ax2 + bx + c.
2. Calculate x coordinate of vertex with the help of formula $x = -$$\frac{b}{2a}$. Plugging x in given equation, find y coordinate of vertex.
3. Find x intercept by putting y = 0 in the equation.
4. Find y intercept by substituting x = 0 in the equation.
5. Now plot the graph of parabola with the help of information so obtained.
If equation is in vertex form:
1. Compare it with vertex parabolic form and determine the vertex (h, k).
2. Find coordinates of focus.
3. Determine the equation of directrix.
4. Now plot vertex, focus, directrix, and other required points and graph the parabola.
Let us consider an example:
### Solved Example
Question: Graph the quadratic equation y2 - 2x - 4 = 0
Solution:
Deduce the given equation in vertex form:
y2 = 2(x + 2)
Comparing it with standard form (y - k)2 = 4a (x - h)
h = -2 and k = 0
So, vertex = (-2, 0)
4a = 2
a = $\frac{1}{2}$
Focus = (h + a, 0) = (-$\frac{3}{2}$, 0)
Equation of Directrix:
x = h - a
x = -$\frac{5}{2}$ = -2.5
Finding the value of Y intercept by substituting x = 0
y2 = 4
y = $\pm$ 2
By using above information, we get the following graph of a parabola: | 1,277 | 4,014 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2018-43 | latest | en | 0.860212 |
Subsets and Splits